Chapter 6 HW Packet FOR THE QUANTITATIVE PROBLEMS FOCUS ON: 1, 4-5, 7, 8, 13, 20, 22, 23, 28, 34, 43, 48 Conceptual Questions 1) What is the correct unit of work expressed in SI units? (16/17) A) kg2 m/s2 B) kg m/s2 C) kg m2/s D) kg m2/s2 The easiest way to remember this so it becomes intuitive is that every time you think of work equaling force times distance you also think of the units—N times m, and then N = kgm/s2, and if you can’t remember that, think of it as F x d = ma x d = (kg)(m/s2)(m)= kgm2/s2. If you keep doing this eventually you won’t even have to think about it—you’ll just know that its kgm2/s2. Answer D (5-10 seconds) 2) Can work be done on a system if there is no motion? (15/17) A) Yes, since motion is only relative. B) No, because of the way work is defined. C) Yes, if an outside force is provided. D) No, since a system which is not moving has no energy. Work is defined as Fd cosθ —if there is no displacement, there is no work. Answer B (10-15 seconds) 3) If you push twice as hard against a stationary brick wall, the amount of work you do (16/17) A) remains constant at zero. B) doubles. C) remains constant but non-zero. D) is cut in half. Work is defined as Fd cosθ --if there is no displacement there is no work. If the wall is stationary, there is no work no matter how hard you push against it. Answer A (10-15 seconds) 4) A 50-N object was lifted 2.0 m vertically and is being held there. How much work is being done in holding the box in this position? (17/17) A) 100 J B) more than 100 J C) less than 100 J, but more than 0 J D) 0 J Work is defined as Fd cosθ --if there is no displacement there is no work. As long as the box is being held in position, no work is being done. Answer D 5) If you walk 5.0 m horizontally forward at a constant velocity carrying a 10-N object, the amount of work you do is (12/17) A) less than 50 J, but more than 0 J. B) equal to 50 J. C) more than 50 J. D) zero. Work is defined as Fd cosθ --therefore, the force must be in the direction of the motion. If you are carrying an object the force on the object is directed upward (at 900) compared to the motion of the object. cos (900) = 0, so Fd cosθ =0, so no work is done. Answer D (10-15 seconds). 6) A container of water is lifted vertically 3.0 m then returned to its original position. If the total weight is 30 N, how much work was done? (17/17) A) No work was done. B) 45 J C) 90 J

D) 180 J In lifting the container, a force equal to the weight of the container is applied over a distance of 3.0 m, and the force is in the direction of the container. So, in lifting the container, (30 N)(3.0 m)(cos(00)=90 J of work is performed on the container. In lowering the container, a force of 30 N must still be applied to the container, and the direction of this force is still in the upward direction. However, now the force is

opposite the direction of the container, so W = Fd cosθ = (30N )(3.0 m)(cos 1800 ) = −90 J . The net work is the sum of the work in both directions, and so = 0J. Answer A (15-30 seconds) 7) Does the centripetal force acting on an object do work on the object? (16/17) A) Yes, since a force acts and the object moves, and work is force times distance. B) Yes, since it takes energy to turn an object. C) No, because the object has constant speed. D) No, because the force and the displacement of the object are perpendicular. Remember that the force causing the displacement has to be in the direction of the displacement. The centripetal force, which would be the force causing an object to travel in a circular path, is perpendicular to the motion of the object traveling in the circle therefore, the angle between the force and the motion would be 900 and the cos of 900 is 0, so Fd cosθ =0. Answer D (15-30 seconds) 8) You throw a ball straight up. Compare the sign of the work done by gravity while the ball goes up with the sign of the work done by gravity while it goes down. (13/17) A) Work is - on the way up and + on the way down. B) Work is - on the way up and - on the way down. C) Work is + on the way up and + on the way down. D) Work is + on the way up and - on the way down. Remember that force due to gravity is always acting in a downward direction. When the ball is traveling upward, this means that the angle between the motion and the direction of force is 1800. As cos 1800 is -1, Fd cosθ will have a negative value. On the way down, the ball is traveling in the same direction as the force so the angle between the direction of the force and the motion is 00, for which cos is 1. Therefore, Fd cosθ will have a positive value. Answer A (15 to 30 seconds) 9) The area under the curve, on a Force versus position (F vs. x) graph, represents (17/17) A) work. B) kinetic energy. C) power. D) potential energy. Finding the area under the curve on a F vs d graph is the same thing as multiplying Fd. Therefore, the area under the curve of this type of graph represents work. Answer A (10 to 15 seconds) 10) On a plot of Force versus position (F vs. x), what represents the work done by the force F? ((17/17) A) the slope of the curve B) the length of the curve C) the area under the curve D) the product of the maximum force times the maximum x See explanation for question 9. Answer C (10-15 seconds) 11) The quantity 1/2 mv2 is (15/17) A) the kinetic energy of the object. B) the potential energy of the object. C) the work done on the object by the force. D) the power supplied to the object by the force. ½ mv2 is the definition for kinetic energy. Be able to derive this from W = F x d. Answer A (10-15 seconds)

12) If the net work done on an object is positive, then the object's kinetic energy (17/17) A) decreases. B) remains the same. C) increases. D) is zero. According to the work energy theorem, the change in kinetic energy equals the amount of work performed on the object. If the net work is positive, this means the final kinetic energy is greater than the initial kinetic energy , which means the kinetic energy must have increased. Answer C (10-15 seconds) 13) If the net work done on an object is negative, then the object's kinetic energy (17/17) A) decreases. B) remains the same. C) increases. D) is zero. According to the work energy theorem, the change in kinetic energy equals the amount of work performed on the object. If the net work is negative, this means the final kinetic energy is less than the initial kinetic energy , which means the kinetic energy must have decreased. Answer A (10-15 seconds) 14) If the net work done on an object is zero, then the object's kinetic energy (13/17)/ A) decreases. B) remains the same. C) increases. D) is zero. According to the work energy theorem, the change in kinetic energy equals the amount of work performed on the object. If the net work is 0, this means the final kinetic energy equals the initial kinetic energy , which means the kinetic energy must have remained the same. Answer B (10-15 seconds) 15) A truck weighs twice as much as a car, and is moving at twice the speed of the car. Which statement is true about the truck's kinetic energy compared to that of the car? (16/17) A) All that can be said is that the truck has more kinetic energy. B) The truck has twice the kinetic energy of the car. C) The truck has 4 times the kinetic energy of the car. D) The truck has 8 times the kinetic energy of the car. As with so many other questions like this the best way to solve this is write out the equation, isolate the

variable you are comparing, and see how changing the other variable must affect it. As KE = 12 mv2

you can see that if you double the mass (which would account for double the weight) and double the speed, the KE must be multiplied by 8 times. Write the formula out if necessary to visualize

KE = 12 mv2 → 1

2 2m 2v2= 8 KE Answer D (30-45 seconds)

16) Car J moves twice as fast as car K, and car J has half the mass of car K. The kinetic energy of car J, compared to car K is (15/17)\] A) the same. B) 2 to 1. C) 4 to 1. D) 1 to 2. Write out the equation, isolate the variable you are comparing, and see how changing the other variable(s) must affect it. Answer D (30-45 seconds)

KE = 1

2 mv2 → 12

12 m 2v

2= 2 KE

17) An object hits a wall and bounces back with half of its original speed. What is the ratio of the final kinetic energy to the initial kinetic energy? (15/17) A) 1/2 B) 1/4

C) 2 D) 4 This problem may be intuitive enough for you at this point that you can readily see the answer should be ¼ of the original KE because KE depends on the square of the velocity. However, it never hurts to write out the equation, isolate the variable you are comparing, and see how changing the other variable(s) must affect it. Answer B (30-45 seconds)

KE = 1

2 mv2 → 12 m 1

2 v2= 1

4 KE

18) A brick is moving at a speed of 3 m/s and a pebble is moving at a speed of 5 m/s. If both objects have the same kinetic energy, what is the ratio of the brick's mass to the rock's mass? (16/17) A) 25 to 9 B) 5 to 3 C) 12.5 to 4.5 D) 3 to 5 In this case write out the formula for KE for both objects, isolate the masses, create the ratio of the masses, then see what applying the velocities does to this ratio. Answer A (60-90 seconds)

KEB = 12 mBvB

2 12 mBvB

2 = 12 mPvP

2 →mB

mP

=vP

2

vB2 →

5P2

3B2 = 25

9

KEP = 12 mPvP

2

19) A 4.0-kg mass is moving with speed 2.0 m/s. A 1.0-kg mass is moving with speed 4.0 m/s. Both objects encounter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping? (16/17) A) the 4.0-kg mass B) the 1.0-kg mass C) Both travel the same distance. D) cannot be determined from the information given Because the same force is used to stop both objects, and ΔKE =W = Fd cosθ , the stopping distance will depend only on the change in kinetic energy—which ever object has the greatest kinetic energy will take the longest to stop. You can quickly determine the first object has a kinetic energy of 8 J, and the second object also has a kinetic energy of 8J. Therefore, both will take the same distance to stop. Answer C (45-60 seconds) 20) You slam on the brakes of your car in a panic, and skid a certain distance on a straight, level road. If you had been traveling twice as fast, what distance would the car have skidded, under the same conditions? A) It would have skidded 4 times farther. B) It would have skidded twice as far. C) It would have skidded 1.4 times farther. D) It is impossible to tell from the information given. This problem may be intuitive enough for you at this point that you can readily see that as the KE is proportional to the square of the velocity, if you double the velocity you will have four times the KE, and because this would equal the amount of work done stopping the car and W = Fd, this would mean four times the stopping distance. However, it never hurts to write out the equation, isolate the variable you are comparing, and see how changing the other variable(s) must affect it. Answer A (10-15 seconds)

KE = 12 m 2v

2=W = F 4d cosθ

21) A planet of constant mass orbits the Sun in an elliptical orbit. Neglecting any friction effects, what happens to the planet's kinetic energy? A) It remains constant. B) It increases continually. C) It decreases continually.

D) It increases when the planet approaches the Sun, and decreases when it moves farther away. As the distance from the sun decreases as the planet approaches the sun, the gravitational force will increase and the velocity of the planet will increase, therefore the KE will increase. As the distance from the sun increases as the planet moves away from the sun, the gravitational force will decrease and the velocity of the planet will decrease, therefore, KE will decrease. Answer D (10-15 seconds) 22) State the work-energy principle. The net work done on an object is equal to the change in the objects kinetic energy. 23) The quantity mgy is A) the kinetic energy of the object. B) the gravitational potential energy of the object. C) the work done on the object by the force. D) the power supplied to the object by the force. We can determine the formula for the gravitational potential energy by recognizing that in increasing an objects gravitational potential, we are doing work against gravity, and W = Fd cosθ . If there is no acceleration as we are raising the object, the force applied to the object must equal the force due to

gravity, but the force is in the opposite direction of gravity so cosθ = cos1800 =-1. If y equals the distance raised, PE=W becomes PE = -m(-g)y=mgy. Answer B (10-15 seconds) 24) The quantity 1/2 kx2 is A) the kinetic energy of the object. B) the elastic potential energy of the object. C) the work done on the object by the force. D) the power supplied to the object by the force. Remember, the force required to stretch a spring a certain distance is given by F = kx , where k is the spring constant (Hooke’s law). To find the work required to stretch the spring, and so, the amount of PE stored in the spring by that amount of work, we cannot just multiply this force times the distance, because the force changes depending on the degree of stretching of the spring. We would therefore need to use the average force, which could be given by

12 (Finitial + Ffinal ) =

12 (0N + Ffinal ) =

12 Ffinal

The expression for PE stored in a spring then becomes PE =W = 12 Fx = ( 1

2 kx)(x) = 12 kx2

Answer B 25) Is it possible for a system to have negative potential energy? A) Yes, as long as the total energy is positive. B) Yes, since the choice of the zero of potential energy is arbitrary. C) No, because the kinetic energy of a system must equal its potential energy. D) No, because this would have no physical meaning. As with all conservative forces, the calculation of potential energy depends on the position of an object with reference to a position representing 0 J of potential energy. If the position of the object is at a position that is negative to this reference, then the calculated potential energy will have a negative value. Answer B (15-30 seconds) 26) An object is released from rest a height h above the ground. A second object with four times the mass of the first if released from the same height. The potential energy of the second object compared to the first is A) one-fourth as much. B) one-half as much. C) twice as much. D) four times as much. Write out the equation, isolate the variable you are comparing, and see how changing the other variable(s) must affect it. Answer D (30-45 seconds)

PE = mgy → 4m gy = 4 PE 27) A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, the potential energy of the spring will be A) the same. B) twice as much. C) 3 times as much. D) 4 times as much. Remember that the F required to stretch a spring equals the spring constant times the distance stretched

( F = kx ). The work required to stretch the spring therefore is W = Fd = (kx)(x) = kx2 --except, F is not constant. As the spring stretches, the force changes so we need to use the average force (force at the beginning and end of the stretch divided by 2= ½ F). This makes the work required to stretch the spring:

W = kx2 . This will also be the amount of PE we add to the spring. Adding another .200 kg of mass to the spring will stretch the spring an additional 5.0 cm. As the potential energy added is multiplied by the square of the distance, the PE will be four times as much. Answer D (45-60 seconds—or less) 28) The total mechanical energy of a system A) is equally divided between kinetic energy and potential energy. B) is either all kinetic energy or all potential energy, at any one instant. C) can never be negative. D) is constant, only if conservative forces act. By definition, the total mechanical energy of a system is defined as KE + PE. If no non-conservative forces are present to modify these energies, the total mechanical energy of a system remains constant. 29) An acorn falls from a tree. Compare its kinetic energy K, to its potential energy U. A) K increases and U decreases. B) K decreases and U decreases. C) K increases and U increases. D) K decreases and U increases. As any object falls under the influence of gravity, potential energy is exchanged for kinetic energy—K increases and U decreases—Answer A (10-15 seconds) 30) Describe the energy of a car driving up a hill. A) entirely kinetic B) entirely potential C) both kinetic and potential D) gravitational As the car has motion it has kinetic energy. As it is traveling up a hill, it is also increasing in potential energy. Answer C (10-15 seconds) 31) A lightweight object and a very heavy object are sliding with equal speeds along a level frictionless surface. They both slide up the same frictionless hill. Which rises to a greater height? A) The heavy object, because it has greater kinetic energy. B) The lightweight object, because it weighs less. C) They both slide to the same height. D) cannot be determined from the information given Although they are both traveling at the same speed, the heavier object will have more KE. However, as they both convert their kinetic energy to potential energy, as mass is in both formulas, it will cancel out and amount of potential energy obtained will only be related to the speed, which is the same for both. So, they will climb to the same height. Answer C (15-30 seconds) 32) Consider two masses m1 and m2 at the top of two frictionless inclined planes. Both masses start from rest at the same height. However, the plane on which m1 sits is at an angle of 30° with the horizontal, while the plane on which m2 sits is at 60°. If the masses are released, which is going faster at the bottom of

its plane? A) m1 B) m2 C) They both are going the same speed. D) cannot be determined without knowing the masses Because the only thing that matters in the conversion of PE to KE in the absence of non-conservative forces is the difference in height, both objects will traveling a the same speed at the end of the incline. Answer C (10-15 seconds) 33) State the principle of conservation of mechanical energy for conservative forces. If only conservative forces do work, the total mechanical energy of a system neither increases nor decreases in any process. It stays constant—it is conserved. 34) A ball falls from the top of a building, through the air (air friction is present), to the ground below. How does the kinetic energy (K) just before striking the ground compare to the potential energy (U) at the top of the building? A) K is equal to U. B) K is greater than U. C) K is less than U. D) It is impossible to tell. If friction was not present, all of the potential energy, U, would be converted to kinetic energy, K, during the fall. However, as air-resistance, a dissipative force, is present, some of the PE will be used to overcome this, and the amount of K present at the ground will be reduced by this amount. Answer C (10-15 seconds) 35) A ball drops some distance and gains 30 J of kinetic energy. Do not ignore air resistance. How much gravitational potential energy did the ball lose? A) more than 30 J B) exactly 30 J C) less than 30 J D) cannot be determined from the information given If we could ignore air resistance, we would say the whole amount of KE the ball gained was originally PE. But, with air resistance, some of the original PE must have been lost. Therefore, the amount of PE the ball must have started with, must have been greater than 30 J. Answer A (10-15 seconds) 36) A ball drops some distance and loses 30 J of gravitational potential energy. Do not ignore air resistance. How much kinetic energy did the ball gain? A) more than 30 J B) exactly 30 J C) less than 30 J D) cannot be determined from the information given If we could ignore air resistance, we would say that the whole amount of PE would be completely transformed into kinetic energy of the ball. However, as we cannot ignore air resistance, the amount of KE the ball gains will be reduced by the work done by air-resistance—so, less than 30 J. Answer C (10-15 seconds) 37) State the law of conservation of energy. The total energy is neither increased nor decreased in any process. Energy can be transformed from one form to another, and transferred from one object to another, but the total amount remains constant. 38) The quantity Fd/t is A) the kinetic energy of the object. B) the potential energy of the object. C) the work done on the object by the force. D) the power supplied to the object by the force. As Fd = W, and so the whole expression is then W/t, this is another version of the definition of power.

Remember that the d/t part of this expression is also v, so P=Fv. Answer D (10-15 seconds) 39) What is the correct unit of power expressed in SI units? A) kg m/s2 B) kg m2/s2 C) kg m2/s3 D) kg2 m/s2

You know that F = ma = kgm / s2

You know that W = ΔKE = Fd = kgm / s2( )(m) = kgm2 / s2( )

P = W

t= kgm2 / s2

s= kgm2 / s3 Answer C (10-15 seconds)

40) Of the following, which is not a unit of power? A) watt/second B) newton-meter/second C) joule/second D) watt The unit of power is given in some form of joules per second—1 J per second is the definition of a watt. Therefore, you know immediately that C and D are units of power. As joules are also newton meters, newton meter per second (letter B) is a unit of power. If watt is a unit of power, then watts per second (letter A) cannot be. Answer A (10-15 seconds) 41) Compared to yesterday, you did 3 times the work in one-third the time. To do so, your power output must have been A) the same as yesterday's power output. B) one-third of yesterday's power output. C) 3 times yesterday's power output. D) 9 times yesterday's power output. Write out the equation, isolate the variable you are comparing, and see how changing the other variable(s) must affect it. Answer D (30-45 seconds)

P = Wt

→ 3W13 t

= 9 P

42) To accelerate your car at a constant acceleration, the car's engine must A) maintain a constant power output. B) develop ever-decreasing power. C) develop ever-increasing power. D) maintain a constant turning speed.

Remember that power equals force times velocity. ( P = W

t= Fd

t= Fv ) If you are constantly

increasing the velocity, you will constantly need to increase P to increase velocity further. Therefore, to accelerate your car at a constant acceleration, the car’s engine must develop ever-increasing power. Answer C (15-30 seconds) Quantitative Problems 1) An object is lifted vertically 2.0 m and held there. If the object weighs 90 N, how much work was done in lifting it? This just requires plugging values into the work formula. A) 360 J B) 180 J

C) 90 J W = Fd cosθ = (90N )(2m)(cos 00 ) = 180 J Answer B

D) 0 J 2) You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity of 0.50 m/s. What is the work done by the weight of the book? Answering this requires recognizing that the work done by A) +10 J the weight of the book is work done by the force due to gravity. As this force is in the B) -10 J opposite direction of the motion, the angle between the force and direction of motion C) +5.0 J is 1800 so the sign of the value of work will be negative. Otherwise, plug the values D) -5.0 J in and solve.

W = Fd cosθ = (10N )(1.0m)(cos 1800 ) = −10 J Answer B

3) A 500-kg elevator is pulled upward with a constant force of 5500 N for a distance of 50.0 m. What is the work done by the 5500 N force? This just requires plugging values into the work formula. A) 2.75 × 105 J

B) -2.45 × 105 J W = Fd cosθ = (5500N )(50 m)(cos00 ) = 275000J = 2.8 x 105 J

C) 3.00 × 104 J Answer A D) -5.20 × 105 J 4) A 500-kg elevator is pulled upward with a constant force of 5500 N for a distance of 50.0 m. What is the work done by the weight of the elevator? Answering this question requires recognizing that the A) 2.75 × 105 J work done by the weight of the elevator is work done by the force due to B) -2.45 × 105 J gravity. As this force is in the opposite direction of the motion, the angle C) 3.00 × 104 J between the force and the direction of motion is 1800 so the sign of the D) -5.20 × 105 J value of work will be negative. Also, you will have to determine the force due to gravity—we are given no indication that the elevator is not accelerating so we cannot assume that the force of gravity is the opposite of the force due to tension in the elevator cable. Why must we use a positive value for acceleration due to gravity in this situation? (because cos 1800 takes care of making the value for work negative in these problems.) Answer B FG = mg = (500 kg)(9.8 m / s) = 4900 N

WG = Fd cosθ = (4900 N )(50.0m)(cos1800 ) = −2.45 x 105 J 5) A 500-kg elevator is pulled upward with a constant force of 5500 N for a distance of 50.0 m. What is the net work done on the elevator? Answering this question requires recognizing that the net work done A) 2.75 × 105 J will equal the net force times the displacement (in the direction of the B) -2.45 × 105 J force). Determine the net force in the standard way and multiply by the C) 3.00 × 104 J distance: D) -5.20 × 105 J

Wnet = Fnetd cosθ = (600 N )(50.0 m)(cos 00 ) = 3.00 x 104 J

Fy = FT + FG = 5500 N + (mg) = 5500 N + (500 kg)(−9.8 m / s2 ) = 600 N∑

Alternatively, as you already know the work performed in both due to tension and gravity (from problems 4 and 5), you can simply sum these to get the net work:

Wnet =WT +WG = 275000 J + (−245000 J ) = 3.00 x 104 J

6) A 30-N box is pulled 6.0 m up along a 37° inclined plane. What is the work done by the weight (gravitational force) of the box? A) - 11 J B) - 1.1 × 102J C) - 1.4 × 102 J D) - 1.8 × 102 J Answering this question requires recognizing that work done by the weight of the box is the same thing as work done by gravity. Also recognize that the force due to gravity is not completely in the direction of the motion, so cosθ will not be 0, 1, or -1. To find the angle recognize that the angle we are wanting is the the angle between the direction of motion and the force. As the direction of motion is 370 up the incline and the direction of the force due to gravity is straight down, this angle is 1270. Otherwise, you are given the force of the box and the distance. Plug the values into the work formula and solve.

W = Fd cosθ = (30N )(6.0m)(cos 1270 ) = −1.1 x 102 J 7) A 4.00-kg box of fruit slides 8.0 m down a ramp, inclined at 30.0° from the horizontal. If the box slides at a constant velocity of 5.00 m/s, what is the work done by the weight of the box? Answering this A) 157 J requires recognizing that work done by the weight of the box is the same B) -157 J + 78.4 J thing as work done by gravity. Also recognize that the force due to C) 78.4 J gravity is not completely in the direction of the motion so cosθ will not D) -78.4 J 0, 1 or -1. To find the angle recognize that the angle we are wanting is the angle between the direction of motion and the force. This is not the same angle that is between FG and FGy, but will be 90 – that angle = 90-30 =60 degrees. From here, just plug the values in the work formula to get the work done by gravity. Remember, the cos term takes care of making work done by gravity either positive or negative so we do not need to put a negative in front of g.

WG = Fd cosθ = (4.00 kg)(9.80 m / s2 )(8.0 m)(cos600 ) = 156.8J = 157 J

8) Matthew pulls his little sister Sarah in a sled on an icy surface (assume no friction), with a force of 60.0 N at an angle of 37.0° upward from the horizontal. If he pulls her a distance of 12.0 m, what is the work done by Matthew? A) 185 J B) 433 J C) 575 J D) 720 J This is a basic application of the work formula. Plug the given values in and solve.

W = Fd cosθ = (60.0N )(12.0m)(cos 37.00 ) = 575 J 13) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final kinetic energy of the cart? A) 1.0 × 103 J B) 2.0 × 103 J C) 2.7 × 103 J D) 4.0 × 103 J As change in KE equals work, knowing that initial KE =0 and being able to find W,

KE final =W = Fd cosθ = (200N )(10m) = 2.0 x 103 J

14) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart? A) 73 m/s B) 36 m/s C) 8.5 m/s D) 6.0 m/s Once we determine the KE as above in question 13, then, knowing that KE =1/2 mv2, we can isolate v

and solve for it. You should probably do the beginning part (#13) of this again because it all fits into one calculation nicely.

KE final =W = Fd cosθ = 12 mv2

v = 2Fd cosθm

= 2(200N )(10m)55kg

= 8.5 m / s

15) A 10-kg mass is moving with a speed of 5.0 m/s. How much work is required to stop the mass? A) 50 J B) 75 J C) 100 J D) 125 J Work required to stop the mass will equal the change in KE from its initial speed to a speed of zero. We have all the values we need to determine change in KE.

W = ΔKE = 0J − 12 mv1

2 = − 12 (10kg)(5.0 m / s)2 = 125 J

16) If it takes 50 m to stop a car initially moving at 25 m/s, what distance is required to stop a car moving at 50 m/s under the same condition? A) 50 m B) 100 m C) 200 m D) 400 m As work equals the change in KE and also equals force required to stop the car times the distance, the stopping distance is directly related to the square of the velocity. If you double the velocity, you will increase the stopping distance by 4 times.

ΔKE =W = Fd cosθ = 0J − 1

2 mv12 = 1

2 mv12 → 1

2 m 2v1

2=F 4d cosθ

17) A spring-driven dart gun propels a 10-g dart. It is cocked by exerting a force of 20 N over a distance of 5.0 cm. With what speed will the dart leave the gun, assuming the spring has negligible mass? A) 10 m/s B) 14 m/s C) 17 m/s D) 20 m/s Given no non-conservative forces, the KE the dart will have upon firing will equal the PE the gun transferred to the dart. The KE can then be equated to its formula and solved for velocity. This is tricky though because we are not given a spring constant—we are specifically given a force applied over a distance and from this we can determine the PE stored, rather than using PE = 1

2 kx2

ΔKE = 12 mv2

2 = ΔPE = Fd cosθ = (20N )(.05m) = 1J

12 mv2

2 = 1J → v = 2(1J )m

= 2J.010kg

= 14 m / s

18) A 100-N force has a horizontal component of 80 N and a vertical component of 60 N. The force is applied to a box which rests on a level frictionless floor. The cart starts from rest, and moves 2.0 m horizontally along the floor. What is the cart's final kinetic energy? A) 200 J B) 160 J C) 120 J D) zero For this problem we are only concerned about the force in the direction of motion, which we have already been given. We can use then work energy principle to say that ΔKE =W = Fd cosθ As the initial KE = 0, ΔKE is KE.

KE =W = Fd cosθ = 80N (2.0m) = 160 J 19) An arrow of mass 20 g is shot horizontally into a bale of hay, striking the hay with a velocity of 60 m/s. It penetrates a depth of 20 cm before stopping. What is the average stopping force acting on the arrow? A) 45 N B) 90 N C) 180 N D) 360 N We know that W = Fd cosθ , so we can find F, if we can find W. We also know that W= ΔKE , so

F = ΔKE

d cosθ. In this case F is working against the direction of motion so cosθ will give us a negative

value.. We also know ΔKE = 0J − 12 mv2 . Putting this all together we get:

F = ΔKE

d cosθ=

0J − 12 mv2

−d=

12 (.020kg)(60m / s)2

.20m= 180N

20) A 15.0-kg object is moved from a height of 7.00 m above a floor to a height of 13.0 m above the floor. What is the change in gravitational potential energy? A) zero B) 1030 J C) 1176 J D) 1910 J Change in PE ( ΔPE ) is simply final PE – initial PE =mgy2 –mgy1: (Note—THE ANSWER LIST IS INCORRECT FOR THIS PROBLEM)

ΔPE = mgy2 − mgy1 = mg( y2 − y1) = (15.0kg)(9.8m / s2 )(13.0m− 7.0m) = 882J 21) A 400-N box is pushed up an inclined plane. The plane is 4.0 m long and rises 2.0 m. If the plane is frictionless, how much work was done by the push? A) 1600 J B) 800 J C) 400 J D) 100 J Remember that in situations where we can ignore non-conservative forces, all work accomplished will be done to change potential energy. The plane is frictionless so the only thing work is doing in this case is increasing the potential energy of the box. Therefore, the work will equal the change in potential energy. As the initial PE could be considered “0,” ΔPE  will  equal  PE2,   so:

W = ΔPE = PE2 − PE1 = mg( y2 − y1) = (400N )(2.0m − 0m) = 800N

22) A 10-kg mass, hung onto a spring, causes the spring to stretch 2.0 cm. What is the spring constant? A) 4.9 × 103 N/m B) 5.0 × 103 N/m C) 20 N/m D) 2.0 N/m According to Hooke’s law, force applied to a spring equals the spring constant times the distance to which the spring is stretch. You are given the information to find the force stretching the spring—the force due to gravity—and the distance. Isolate the spring constant from this formula, plug the values in and solve.

FG = kx → k =

FG

x= (10 kg)(9.8 m / s2 )

.02 m= 4.9 x 103 N / m

23) A spring is characterized by a spring constant of 60 N/m. How much potential energy does it store,

when stretched by 1.0 cm? A) 3.0 × 10-3 J B) 0.30 J C) 60 J D) 600 J You know that the PE stored in a stretched spring is given by

PE = 12 kx2 = 1

2 (60 Nm )(.01m)2 = 3.0 x 10−3 J

24) Calculate the work required to compress an initially uncompressed spring with a spring constant of 25 N/m by 10 cm. A) 0.10 J B) 0.13 J C) 0.17 J D) 0.25 J You know that the amount of work required to store energy in a spring equals the change in potential of the spring during the work which equals

12 kx2 . So, W = ΔPE = 1

2 kx2 . Plug values in and solve.

W = ΔPE = 12 kx2 = 1

2 (25 Nm )(.10m)2 = .13 J

25) What work is required to stretch a spring of spring constant 40 N/m from x = 0.20 m to 0.25 m? (Assume the unstretched position is at x = 0.) A) 0.45 J B) 0.80 J C) 1.3 J D) 0.050 J The work required to change the spring from one state to another will equal the corresponding change in PE. So, we need to determine the PE at each state and subtract the initial from the final state.

Remember that the potential energy stored at any state of stretch or compression is 12 kx2 .

W = ΔPE = 12 kx2

2 − 12 kx1

2 = 12 (40 N

m )(.25m2 − .20m2 ) = .45J 26) A spring with a spring constant of 15 N/m is initially compressed by 3.0 cm. How much work is required to compress the spring an additional 4.0 cm? A) 0.0068 J B) 0.012 J C) 0.024 J D) 0.030 J Again, you have a spring compressed at an initial state (3.0 cm) and at a final state (7.0 cm). The additional work required to compress the spring will be the difference between the two states—very similar to question 25. So, we need to determine the PE at each state and subtract the initial from the

final state. Remember that the potential energy stored at any state of stretch or compression is 12 kx2 .

W = ΔPE = 12 kx2

2 − 12 kx1

2 = 12 (15 N

m )(.07m2 − .03m2 ) = .030J 27) A 60-kg skier starts from rest from the top of a 50-m high slope. What is the speed of the skier on reaching the bottom of the slope? (Neglect friction.) A) 22 m/s B) 31 m/s C) 9.8 m/s D) 41 m/s This will be a basic application of the conservation of energy. You know that the initial KE term and final PE terms are not necessary so you can simply equate the final kinetic energy term with the initial PE term and solve for v.

KE2 − PE2 = KE1 − PE1

12 mv2

2 = m gy1 → v = 2gy1 = 2(9.8m / s2 )(50m) = 31 m / s

28) A skier, of mass 40 kg, pushes off the top of a hill with an initial speed of 4.0 m/s. Neglecting friction, how fast will she be moving after dropping 10 m in elevation? A) 7.3 m/s B) 15 m/s C) 49 m/s D) 196 m/s This will be a basic application of the conservation of energy. However, you can only leave out the final PE term in this case.

KE2 − PE2 = KE1 − PE1

12 mv2

2+ = 12 mv2

1 + mgy1 → v2 = v12 + 2gy1 = (4.0m / s)2 + 2(9.8m / s2 )(10m) = 15m / s

29) An object slides down a frictionless inclined plane. At the bottom, it has a speed of 9.80 m/s. What is the vertical height of the plane? A) 19.6 m B) 9.80 m C) 4.90 m D) 2.45 m As we are told we can ignore non-conservative forces, this is a basic conservation of mechanical energy problem. We use KE and PE terms for the before and after circumstance. The initial KE term and the final PE term drop out. As these problems become more intuitive, it is not necessary to include all of the terms, only those you are going to need. For this problem you recognize that the final kinetic energy should equal the initial potential energy—just start with those two terms and isolate y.

KE2 − PE2 = KE1 − PE1

12 mv2

2 = m gy1 → y1 =12 v2

2

g=

12 (9.80m / s)2

(9.8m / s)= 4.90m

30) A 1.0-kg ball falls to the floor. When it is 0.70 m above the floor, its potential energy exactly equals its kinetic energy. How fast is it moving? A) 3.7 m/s B) 6.9 m/s C) 14 m/s D) 45 m/s At .70 m above a floor, a 1.0 kg ball has mgy = (1.0 kg)(9.8 m/s2)(.70m)=6.86J of PE. We are given that it must therefore also have 6.86 J of KE. We can use the KE formula to isolate and determine v.

PE = mgy = (1.0 kg)(9.8m / s2 )(.70 m) = 6.86 J

KE = 12 mv2 → v = 2KE

m= 2(6.86J )

1.0kg= 3.7 m / s

31) A toy rocket, weighing 10 N, blasts straight up from ground level with a kinetic energy of 40 J. At the exact top of its trajectory, its total mechanical energy is 140 J. To what vertical height does it rise? A) 1.0 m B) 10 m C) 14 m D) 24 m

At the top of its trajectory, if fired straight up the total mechanical energy is entirely PE. If it started with 40 J of KE, that 40 J of KE is now converted into PE, plus, 100 J of PE was added to the system through chemical energy transformed into mechanical energy. The 140 J of PE = mgy. Isolate y and solve.

mgy = 140J → y = 140J

mg= 140J

10N= 14 m

32) A projectile of mass m leaves the ground with a kinetic energy of 220 J. At the highest point in its trajectory, its kinetic energy is 120 J. To what vertical height, relative to its launch point, did it rise? A) 220/(mg) meters B) 120/(mg) meters C) 100/(mg) meters D) Impossible to determine without knowing the angle of launch If the projectile leaves the ground with KE of 220 J and at its greatest height still has a KE 120 KE, it must have converted 100 J of KE into 100 J of PE, and with the 120 J of KE it still has moving in the horizontal direction—that is, at the top of a parabolic trajectory. Setting 100 J equal to PE=mgy, y=PE/mg=100/(mg) meters. Answer C (30-60 seconds). 33) A roller coaster starts from rest at a point 45 m above the bottom of a dip Neglect friction, what will be the speed of the roller coaster at the top of the next slope, which is 30 m above the bottom of the dip? A) 14 m/s B) 17 m/s C) 24 m/s D) 30 m/s As we are told we can ignore non-conservative forces, this is a basic conservation of mechanical energy problem. We use KE and PE terms for the before and after circumstance. If we set our final point at 30 m above the bottom of our dip equal to our “0” point, we can let the initial point be at 15 m, and we can drop out the second gravitational PE term and the first KE term. . Isolate v2, plug values in and solve.

12 mv2

2 + mgy2 = 12 mv1

2 + m gy1

v2 = 2gy1 = 2(9.8m / s2 )(15m) = 17 m / s

34) A roller coaster starts with a speed of 5.0 m/s at a point 45 m above the bottom of a dip Neglect friction, what will be the speed of the roller coaster at the top of the next slope, which is 30 m above the bottom of the dip? A) 12 m/s B) 14 m/s C) 16 m/s D) 18 m/s As we are told we can ignore non-conservative forces, this is a basic conservation of mechanical energy problem. We use KE and PE terms for the before and after circumstance. If we set our final point at 30 m above the bottom of our dip equal to our “0” point, we can let the initial point be at 15 m, and we can drop out the second gravitational PE term. However, we will have to maintain both KE terms. Isolate v2, plug values in and solve.

12 mv2

2 + mgy2 = 12 mv1

2 + m gy1

v2 = v12 + 2gy1 = (5.0m / s)2 + 2(9.8m / s2 )(15m) = 18 m / s

35) A roller coaster starts at a point 30 m above the bottom of a dip with a speed of 25 m/s Neglect friction, what will be the speed of the roller coaster at the top of the next slope, which is 45 m above the bottom of the dip? A) 14 m/s B) 16 m/s C) 18 m/s

D) 20 m/s As we are told we can ignore non-conservative forces, this is a basic conservation of mechanical energy problem. We use KE and PE terms for the before and after circumstance. If we set our initial point at 30 m above the bottom of our dip equal our “0” point, we can let the final point be at 15 m, and we can drop out the first gravitational PE term. However, we will have to maintain both KE terms. Isolate v2, plug values in and solve.

12 mv2

2 + m gy2 =12 mv1

2 + mgy1

v2 = v12 − 2gy2 = (25m / s)2 − 2(9.8m / s2 )(15m) = 18 m / s

37) A pendulum of length 50 cm is pulled 30 cm away from the vertical axis and released from rest. What will be its speed at the bottom of its swing? A) 0.50 m/s B) 0.79 m/s C) 1.2 m/s D) 1.4 m/s Recognize that the only thing that matters in determining the speed at the bottom of the swing (the KE at all of the PE the pendulum has when it is pulled away from the vertical, is converted completely to KE at the bottom of the swing. If we can determine the amount of PE it had, we can determine the amount of KE it obtains, and so, its velocity. If we let the y position at the bottom of the swing equal 0 (y2=0), we can determine the vertical difference in height of the pendulum in its two positions by analyzing the triangle formed by the length of the pendulum at its two positions—see diagram. We can cut a little time out of this if we recognize the triangle is a 3-4-5 triangle—the unknown side must be .4 m, so the vertical height must be .1m. (In fact, this is a conservation of energy perspective).

ΔKE = −ΔPE → 12 mv2

2 + mgy2 = 12 mv1

2 + m gy1

v2 = 2gy1 = 2(9.8m / s2 )(.1m) = 1.4 m / s

38) A 1500-kg car moving at 25 m/s hits an initially uncompressed horizontal spring with spring constant of 2.0 × 106 N/m. What is the maximum compression of the spring? (Neglect the mass of the spring.) A) 0.17 m B) 0.34 m C) 0.51 m D) 0.68 m The amount of work done compressing the spring is equal to the amount of work done by the spring in changing the KE of the car, which equals the change in KE energy of the car. As the change of KE will be a negative value, the force applied by the spring will be in a negative direction, and the force of the car compressing the spring will be in a positive direction.

change in potential energy experienced by the spring, and is given by Wapplied = PEel =

12 kx2 . You are

given the information to find the amount of work used in compressing the spring. Isolate x, plug in the values and solve.

Wapplied to spring =12 kx2 → x =

2(Wapplied )k

= 2(468750 J )2.0 x 106 N / m

= .68 m

Wapplied to spring = −Wapplied to car = ΔKE = 0J − 12 mv1

2 = − 12 (1500 kg)(25 m / s)2 = −468750J

Wapplied to spring = 468750 J

39) A driver, traveling at 22 m/s, slows down her 2000 kg car to stop for a red light. What work is done by the friction force against the wheels?

A) -2.2 × 104 J B) -4.4 × 104 J C) -4.84 × 105 J D) -9.68 × 105 J Recognize that there is no applied force opposing the force of friction—friction is the only force acting on this object—so, the net force is just the force due to friction. There is also no PE component—we can approach this problem from a work-energy perspective rather than a conservation of energy perspective—that is: Wnet = ΔKE = Fnetd cosθ . In this problem you are given the information to find the change in KE so all you need to do is calculate this to find the work done by friction. You do not need to worry about force or distance in this problem. Obviously I and the answer list have a difference of opinion about how many SF there are here.

W = ΔKE = 0J − 12 mv1

2 = − 12 (2000 kg)(22 m / s)2 = −5 x 105 J

40) The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25.0 m/s, what distance will it slide before coming to a stop? A) 15.0 m B) 30.0 m C) 375 m D) 750 m Recognize that there is no applied force opposing the force of friction—friction is the only force acting on this object—so, the net force is just the force due to friction. There is also no PE component—we can approach this problem from a work-energy perspective rather than a conservation of energy perspective—that is: Wnet = ΔKE = Fnetd cosθ . Isolate d. You are already given the friction force and you can determine the initial KE. You know that final KE is 0J.

Wnet = ΔKE = Fnetd cosθ → d = ΔKE

Ffr cosθ=

0J − 12 mv1

2

(50.0N )(cos1800 )=−( 1

2 (60.0 kg)(25.0 m / s)2

−50.0N= 375 m

41) A 12-kg object is moving on a rough, level surface. It has 24 J of kinetic energy. The friction force on it is a constant 0.50 N. How far will it slide? A) 2.0 m B) 12 m C) 24 m D) 48 m Recognize that although the object is starting with 24 J of KE, there is no applied force opposing the force of friction—friction is the only force acting on this object—so, the net force is just the force due to friction. There is also no PE component—we can approach this problem from a work-energy perspective rather than a conservation of energy perspective—that is: Wnet = ΔKE = Fnetd cosθ . Also, notice that the change in KE will be negative. All we need to do is isolate d, plug in the values and solve. Recognize that Ffr is in the opposite direction of the motion so it is cos 1800.

Wnet = ΔKE = Fd cosθ → d = ΔKE

F cosθ= 0J − 24J

(.50N )(cos1800 )= −24J−.50N

= 48m

42) A force of 10 N is applied horizontally to a 2.0-kg mass on a level surface. The coefficient of kinetic friction between the mass and the surface is 0.20. If the mass is moved a distance of 10 m, what is the change in its kinetic energy? A) 20 J B) 39 J

C) 46 J D) 61 J As there is no potential energy component here you can work this problem from a work-energy perspective-- ΔKE =W = Fd cosθ --but you will have to determine a net force. You are given an applied force and the ability to determine the opposing Ffr. Find the net force and multiply times the distance to get the change in KE.

ΔKE =Wnet = Fnetd cosθ = (6.08N )(10.0m) = 61 J

F = FA + Ffr = 10N + (−3.92N ) = 6.08N∑Ffr = µmg = (.20)(2.0kg)(−9.8 m / s2 ) = −3.92 N

2 43) A 60-kg skier starts from rest from the top of a 50-m high slope. If the work done by friction is -6.0 × 103 J, what is the speed of the skier on reaching the bottom of the slope? A) 17 m/s B) 24 m/s C) 34 m/s D) 31 m/s This energy analysis includes a non-conservative force (friction) so we need to use the conservation of energy approach. We use KE and PE terms for the before and after circumstance and either subtract work done by friction from the before circumstance or add friction to the after circumstance. The terms for KE in the before circumstance and PE in the after circumstance drop out (use the final position of the box a “zero”). Isolate v2. Work done by friction has already been given to you so you do not need to calculate this. THE ANSWER LIST IS INCORRECT ON THIS PROBLEM!!

12 mv2

2 + mgy2 = 12 mv1

2 + mgy1 − FFrd

v2 =2(mgy1 −Wfr )

m= 2((60kg)(9.8m / s2 )(50 m)− (6000 J ))

60 kg= 28 m / s

44) An 800-N box is pushed up an inclined plane. The plane is 4.0 m long and rises 2.0 m. It requires 3200 J of work to get the box to the top of the plane. What was the magnitude of the average friction force on the box? A) 0 N B) non-zero, but less than 400 N C) 400 N D) greater than 400 N This is more of a conceptual problem than a quantitative one. We are not given information to the contrary so we can assume there is no acceleration. Drawing a diagram will help in this circumstance to see that the FA up the incline is opposed by both the Ffr and the x component of FG. Also, the length (hypotenuse) and height (opposite) of the incline show us that the sin of the angle of incline must be .5, which you should know means that the angle is 300. Because FGx = mg sinθ we can calculate the x component of the weight (this should be considered a negative value as it is in the direction opposite to that of the applied force). Also, as we are given the work to push the box up the incline and distance up the incline, we can determine FA. Subtracting FGx from FA will give us Ffr. (The way I have done this gives us a negative value for the force due to friction—the answer list obviously just wants the absolute value of the force due to friction. You could probably reason this out in your head without doing the calculations.

Fx = FA + FGx+ Ffr∑ = 0N → Ffr = −FA − FGx = −800N − (−400N ) = −400N

FGx = mg sinθ = (800 N )(.5) = 400 N

W = FAd → FA =W / d = 3200J / 4.0 m = 800 N

45) A 2.0-kg mass is released from rest at the top of a plane inclined at 20° above horizontal. The coefficient of kinetic friction between the mass and the plane is 0.20. What will be the speed of the mass after sliding 4.0 m along the plane? A) 2.2 m/s B) 3.0 m/s C) 3.5 m/s D) 5.2 m/s This energy analysis includes a non-conservative force (friction) so we need to use the conservation of energy approach. We use KE and PE terms for the before and after circumstance and either subtract work done by friction from the before circumstance or add friction to the after circumstance. The terms for KE in the before circumstance and PE in the after circumstance drop out (use the final position of the box a “zero”). Isolate v2. To find y1 you will have to calculate it knowing the angle of the incline, taking the distance sliding down the incline (4.0 m) as the hypotenuse—the height of this triangle will be y1. To find Ffr, you know that it will equal µFN and FN is the opposite of the y-component of weight.

12 mv2

2 + mgy2 = 12 mv1

2 + mgy1 − FFrd

v2 =2(mgy1 − FFrd)

m= 2((2)(9.8m / s2 )(1.368 m)− (3.68 N )(4.0m)

2.0 kg= 3.5 m / s

sinθ = y1hyp → y1 = sin200( )(4.0m) = 1.368 m

Ffr = µFN = µFGy = µmg cosθ = (.20)(2.0kg)(9.8 m / s2 )(cos200 ) = 3.68 N

46) A 30.0-N stone is dropped from a height of 10.0 m, and strikes the ground with a velocity of 7.00 m/s. What average force of air friction acts on it as it falls? A) 22.5 N B) 75.0 N C) 225 N D) 293 N This energy analysis includes a non-conservative force (air resistance) so we need to use the conservation of energy approach. We use KE and PE terms for the before and after circumstance and either subtract work done by air resistance from the before circumstance or add air resistance to the after circumstance. The terms for KE in the before circumstance and PE in the after circumstance drop out. Isolate FAR, plug in values and solve (need to convert the weight of the stone to mass for the remaining KE term)

12 mv2

2 + mgy2 = 12 mv1

2 + mgy1 − FARd

FAR =mgy1 − 1

2 mv22

d=

(30.0N )(10.0m)− 12 ( 30.0 N

9.8m/s2 )((7.0m / s)2

10.0m= 22.5 N

47) At what rate is a 60.0-kg boy using energy when he runs up a flight of stairs 10.0-m high, in 8.00 s? A) 75.0 W B) 735 W C) 4.80 kW D) 48 W Power equals work per unit of time. Work in this case is the change in PE as a result of the KE used to climb the stairs (this would be a case where you could just use the work-energy concept instead of the conservation of energy) W = ΔPE = ΔKE . So, work equals mgy in this case. Calculate this and divide by time.

P = W

t= mgy

t= (60.0 kg)(9.8 m / s2 )(10.0 m)

8.00 s= 735W

48) A 10-N force is needed to move an object with a constant velocity of 5.0 m/s. What power must be delivered to the object by the force? A) 0.50 W B) 1.0 W C) 50 W D) 100 W

Remember that it is useful to know that P = Fv (derived by P = W

t= Fd

t= Fv ). Plug the values in

and solve. P = Fv = (10 N )(5.0 m / s) = 50W 49) How many joules of energy are used by a 1.0 hp motor that runs for 1.0 hr? (1 hp = 746 W) A) 3.6 × 103 J B) 2.7 × 106 J C) 4.5 × 104 J D) 4.8 J As power is in J/s to find energy, multiply W times number of seconds (3600 s/hr) Even though this is the same thing as J, the unit of energy after this multiplication could also be termed watt seconds ( w ⋅s ) (Note: although you would think it would be easier just to quantify energy in terms of J, convention in the electricity industry is to use the term kilowatt hours (NOT kW per h, but kWh).

1.0 hp( ) 746W

hp⎛⎝⎜

⎞⎠⎟

3600s( ) = 2.7 x 106 J

50) A cyclist does work at the rate of 500 W while riding. How much force does her foot push with when she is traveling at 8.0 m/s? A) 31 N B) 63 N C) 80 N D) 4000 N

For vehicles remember that it is useful to know that P = Fv (derived by P = W

t= Fd

t= Fv ). Isolate

force, plug in the values and solve. P = Fv → F = P

v= 500W

8.0 m / s= 63 N

51) A 1500-kg car accelerates from 0 to 25 m/s in 7.0 s. What is the average power delivered by the engine? (1 hp = 746 W) A) 60 hp B) 70 hp C) 80 hp D) 90 hp

For vehicles remember that it is useful to know that P = Fv (derived by P = W

t= Fd

t= Fv )You also

know that F = ma and you have been given information to determine acceleration. Velocity would have to be average velocity which you could find by taking dividing the sum of the initial and final velocity by 2. You also need to convert W to hp.

P = Fv = mav = 1500 kg( ) 25m / s− 0m / s

7.0s⎛⎝⎜

⎞⎠⎟

25m / s+ 0m / s2

⎛⎝⎜

⎞⎠⎟= 66964W x

1 hp746W

= 90. hp