set of exxperiment

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Experiment No. 4NEWTONS SECOND LAW OF MOTION

Objective:To study Newtons second law and to show that the acceleration of a body is directly proportional to the accelerating force and inversely proportional to the mass of the body.

Apparatus:smooth running single pulley, set of weights, meter stick, light string, stopwatch and iron stand or force frame.

Theory:

Isaac Newton formulated three principles based upon observations he and others had made which summarize the behavior of moving bodies that they have become known as the laws of motion. The three laws of motion are as follows:

First Law of Motion (The Law of Inertia)An object at rest will remain at rest and an object in motion will continue in motion in a straight line at constant speed in the absence of any interaction with the rest of the universe.

Inertia is the property of matter to resist any change in its state of rest or uniform motion. The mass of an object determines its inertia.

Second Law of Motion (The Law of Acceleration)The net force or an unbalanced force acting upon an object will produce an acceleration directly proportional and of the same direction as the force but inversely proportional to the mass of the body.A force is any influence that can change the speed or direction of motion of an object. Weight is a force that causes the object to be accelerated when it is dropped. The mass of a body is the same everywhere unlike its weight, which depends on its position relative to the earth, or to other astronomical body.

The first law describes motion of objects when net force is zero. The second law states that acceleration occurs when a net force is acting. These two laws do not tell the origin of the forces.

Third Law of Motion (The Law of Interaction)When an object exerts a force on another object, the second object exerts on the first a force of the same magnitude but in the opposite direction.

The third law simply states that every action force is accompanied by a reaction force. The two forces act on different bodies.

Procedures:1. Mount a smooth pulley of negligible mass on a force frame. 2. Connect two weights of 20 g each by means of a light string that passes over a frictionless pulley that is mounted on the force frame. Make the distance of the two weights about 100 cm from the foot of the frame.3. Add 5 g to 10 g weight to one of the masses attached to the string making it heavier than the other mass. Determine the time it would take for the body to descend to the foot of the frame. Repeat this procedure several times and get the average time.4. Using a meter stick, measure the distance the body has descended. Determine the acceleration of the system using the equation , knowing that with v1 = 0.5. Repeat steps 3 and 4 using another weights to represent the heavier body while maintaining the 20 g weight as the lighter body.

PAGE 180Data and Observation:

Trial 1Trial 2

Mass of heavier body, M_________ gm_________ gm

Mass of lighter body, m_________ gm_________ gm

Height of fall, S_________ cm_________ cm

Average time of descent, t(Average of 5 trials)_________ sec_________ sec

Acceleration, aExperimental value, Accepted value, _________

_________ _________

_________

Percent of error_________ %_________ %

Computation and Analysis:1. Compute the experimental value of the acceleration, a, using the equation , where s is the average descent of the heavier body for 5 trials, and t is the corresponding average time.2. Using Newtons second law of motion, derive the equation for the theoretical value of a which is given by the equation . Conclusion:

Review Questions:

1. State Newtons second law of motion.2. Differentiate mass and weight. Is your mass on the earth the same when you are on another planet? What about your weight? Explain your answer.3. Compare the accelerating force and the acceleration of trial 1 and trial 2 of your experiment.4. What is the unit of mass and weight in the English system of units?5. Based on your experiment, discuss the principle of the motion of an elevator.


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Experiment No. 5FRICTION

Objectives:1. To study frictional force.

2. To determine and measure the coefficients of static and kinetic friction.

Apparatus:friction block, string, set of slotted masses, weight holder, platform balance, meter stick.

Theory:Friction is a common force that resists the sliding surface through another surface due to interlocking of projections and depressions of surfaces in control. When one body moves another body, frictional force acts parallel to the surfaces in contact and opposite to the motion of the bodies.

Friction depends upon the nature of surfaces in contact. It is directly proportional to the normal force pressing the surfaces together. Regardless of the area in contact, the greater the normal force, the greater the frictional force. In symbols:

Nf = Nf

hence the coefficient of friction is the ratio of the frictional force to the normal force.

where:f = frictional force, N = normal force, and = coefficient of friction.

Friction acting between bodies at rest is static friction (fs) while friction between bodies that are in motion is kinetic friction (fk).

The constant of proportionality s between fs and Nf is called the coefficient of static friction.

When the friction block has started to move, a smaller applied force F is needed to keep the block moving in a uniform motion. This force is opposed by the kinetic friction fk which is given by the formula:

Procedure:A. Measurement of coefficient of static friction1. Weigh the friction block2. Place the block on the friction board and attach a string that, passes over the pulley to the weight holders with slotted masses.3. Determine the force F necessary just to impend the block from moving.4. Repeat using different loads on the block.5. Calculate the coefficient of static friction.B. Measurement of coefficient of kinetic friction1. Repeat procedures A-1 and A-2.2. Put sufficient load on the weigh pan so that when the block is given a light push, it will slide slowly without acceleration. Record the weight of the weight pan and the load on it. Make three trials with different loads on the block.3. Repeat procedures B-1 and B-2 with different areas (smaller and larger portion) of the friction block in contact with friction board.4. Determine the coefficient of kinetic friction.C. Inclined Plane1. Remove the string from the block and place the block near the top of the wooden plane. Elevate this end gradually until the block slides down the plane at constant speed. Get the angle of inclination of the board by measuring the length and the base of the plane. Make three trials by adding load on the block.2. Compute the coefficient of friction using the formula .

Data and Observation

A. Measurement of coefficient of static friction, sTrialsWt. Of block + Wt. Added = NfWt. Of Pan + Wt. Added = fsCoefficient of static friction, s

1

2

3

Ave.

.

B. Measurement of coefficient of kinetic friction, kB-ITrialsWt. Of block + Wt. Added = NfWt. Of Pan + Wt. Added = fsCoefficient of kinetic friction, k

1

2

3

Ave.

B-IIUsing different areas of the friction blockTrialsAreaWt. Of block + Wt. Added = NfWt. Of Pan + Wt. Added = fsCoefficient of kinetic friction, s

1Smaller

2

Ave.

1Larger

2

Ave.

C. Inclined PlaneTrialshb

Weight of block =

Wt of block + wt added =

.

Computation and Analysis:1. Using the data of static friction, construct a graph, the frictional force as the ordinate and the normal force as the abscissa. What does the slope represent?2. Compare the average values of s and k. Which is greater? Why?3. From the data of part B-II, determine the coefficient of kinetic friction for different areas of contact. Does the frictional force depend on the area of contact?4. Compute the coefficient of static friction using the data in part C.5. How do you compare the values of coefficient of static friction as obtained in horizontal plane (part A) and inclined plane (part C).

Conclusion:

Review Questions:1. Name several types of mechanisms in which friction is essential for proper operation.2. What becomes of the energy expended against friction?3. Why should one take short steps rather than long ones when walking on ice?4. Differentiate static friction from kinetic friction.5. What is meant by normal force? Is it always equal to the weight? Explain.


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Experiment No. 6TORQUE

Objective:To study the concept of torque and the conditions that must be satisfied to attain rotational equilibrium.

Apparatus:meter stick, knife-edge support (or force frame), set of weights, weight hangers/holders, platform balance, spring balance

Theory:

Torque (or moment) is a measure of the tendency of a force to cause an object to turn or rotate. The effect of a force producing rotation depends upon the magnitude and direction of the force and location of the axis of rotation. The torque or the moment of force about any axis is the product of the force and the perpendicular distance from the axis to the line of action of the force. The defining equation is = Fswhere: is the torque developed by the force which acts at a perpendicular distance s from the axis. This distance is frequently referred to as the lever arm.

A body that has no translational or rotational acceleration is said to be in equilibrium. Two conditions must be met in order for equilibrium to be attained with several parallel forces. They are: the sum of the forces acting downward must equal the sum of the forces acting upward; and the sum of the clockwise torques must equal the sum of the counterclockwise torques.In rotating a body about some axis, it is apparent that the problem in locating the center of gravity and the center of mass arises. The center of gravity is the point through which the action line of the weight always passes while the center of mass of an object is the single point that moves in the same way as a point mass would move when subjected to the same external forces that act on the object.

The algebraic sign of a torque is perfectly arbitrary. For convenience, clockwise torques are usually called positive and counterclockwise torques negative.

Procedures:1. Set-up the apparatus and locate the center of gravity of the meter stick by adjusting the point of suspension until the meter stick is in equilibrium. Record this position as the center of gravity of the meter stick.2. At the 10 cm mark of the meter stick, hang a 40 g mass. If using a holder or hanger, be sure to include its mass to the given load. Locate the position at which an 80 g be hung to produce rotational equilibrium.3. After equilibrium has been attained, convert the masses into force equivalents and the lever arms into meter. Record all torque producing forces and their respective lever arms (moment arms). Identify each torque as clockwise or counterclockwise.4. Attach two weights on the meter stick: at 10 cm mark, hang a 20 g mass and at 20 cm mark, 10 g mass. Balance their moments by a 60 g mass hung on the other side of the fulcrum. Repeat procedure 3 and compute the percentage deviation.5. Slide the point of suspension of the meter stick as far as possible toward the zero end of the meter stick. Record this position as the axis of rotation. With a loop of thread and a spring balance, apply a force near the 100 cm end of the meter stick so that equilibrium is produced. Record the force exerted by the spring balance and the moment arm. From the law of moments, calculate the weight of the meter stick. Then weigh the meter stick on the triple beam balance and record the relative deviation between the values.


Center of gravity of the meter stick = cm mark

Step 3Massm(kg)ForceF(Newton)Lever ArmS(m)Moment of Force(Newton - meter)

+-

+-

Percentage Difference = %

Step 4Massm(kg)ForceF(Newton)Lever ArmS(m)Moment of Force(Newton - meter)

+-

+-

Percentage Difference = %

Weight of the meter stick using:a. spring balance = Nb. beam balance = Nc.Percentage Difference = %

Computation and Analysis:

Conclusion:

Review Questions:1. Define the following terms:a. torque or moment

b. lever arm or moment arm

c. equilibrium

d. center of gravity

e. center of mass

2. What two conditions must be satisfied for an object to be in equilibrium?3. Describe ways by which an experimenter could locate the center of gravity of irregularly shaped objects.4. Three unequal forces act upon a body at a point so that the body is in equilibrium. If the magnitudes of two of the forces are doubled, how must the third force be changed to preserve equilibrium? Justify your conclusion by diagrams. 5. What is the relationship between the terms axis of rotation and center of gravity?


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Experiment No. 7SIMPLE MACHINES

Objectives:1. To calculate the mechanical advantages of an inclined plane, pulley systems, and wheel and axle.

2. To determine the efficiency of the different types of simple machines.

Apparatus:inclined plane, slotted masses, wheel and axle, triple beam balance, weight holder, double three-pulley system, block of wood, string.

Theory:

Energy is the capacity to do work. To perform work, machines must receive energy from different sources and the maximum work they do must exceed the energy they receive.

Machines can be used to multiply force at the expense of speed or vice versa. However, it does not multiply work input as proven by the Law of Conservation of Energy.

In a simple machine, the energy is supplied by a single applied force which does useful work against a single resisting force.

Most machines, no matter how complex they are, are combinations of two or more simple machines. There are six types of simple machines that are classified into two groups such as: the lever group and the inclined plane group. Under the lever group is the lever itself, the pulley and the wheel and axle. While in the inclined plane group are the inclined plane, the screw and the wedge.

Work, as performed by machines, is the product of the force and its displacement in the direction of the force.

The ratio of the force exerted to the machine or input force over the force exerted by the machine or output force is called actual mechanical advantage (AMA).Mathematically,

where:Fo = force output and Fi = force input.

Likewise, the ratio of the distance Si, through which the input force acts to the distance So, through which the output force acts is the ideal or theoretical mechanical advantage.

That is,

On the other hand, the ratio of the output work, FoSo to the work input FiSi multiplied by 100% is the efficiency of the machine. Similarly, the ratio of the AMA over the IMA multiplied by 100% is also equal to the efficiency.

Efficiency is always expressed in percent and it is always less than 100% due to friction.

Another way of computing the efficiency is by using the formula

Procedures:1. Inclined Planea. Adjust the inclined plane until it is at 10 with the horizontal.b. Measure the length and height of the inclined plane.c. Record the weight of a block of wood and add suitable loads on top of it.d. Get a piece of string. Tie one end on the loaded block of wood and the other end on the weight holder. e. Place the loaded block of wood on the bottom of the inclined plane and hang the other end of the string tied to the other holder on the frictionless pulley connected to the upper end of the inclined plane.f. Add weights on the holder until it is just enough to pull the loaded block up the plane with constant speed.g. Repeat steps a to f using 30 and 40 angles of inclination.2. Wheel and Axlea. Measure the radius of the wheel and also of the axle.b. Wind a string several times around the axle and another string several times around the smaller wheel in the reverse direction.c. Hang a 200 g load at the end of the string coming from the axle. Hang a smaller or enough weight at the end of the string coming from the wheel and gradually increase the weight until the 200 g load moves upward at a constant speed.d. Repeat the above steps using bigger wheels.3. Pulley Systema. Set up the apparatus by winding the string around the two-pulley system.b. Place a 200 g mass on the weight holder that is suspended on the upper end of the system, hang a suitable weight that will make the pulley system move up or down with the same velocity once it has been started.c. Repeat the above steps using a three-pulley system and a five-pulley system.

Data and Observations

A. Inclined PlaneAngle of Inclination

100300400

Weight of block and load (dyne)

Input force (dyne)

Length of inclined plane (cm)

Height of inclined plane (cm)

AMA

IMA

Efficiency (100%)

.

B. Wheel and AxleNumber of Wheels

123

Weight of load (dyne)

Input force (dyne)

Radius of wheel

Radius of axle

AMA

IMA

Efficiency (100%)

.

C. Pulley SystemNumber of Pulleys

DoubleTriple5 Pulley System

Weight of load (dyne)

Input force (dyne)

Output distance

Input distance

AMA

IMA

Efficiency (100%)

.

Computation and Analysis:1. Compute the mechanical advantage of the inclined plane, wheel and axle and set of pulleys used.2. Determine the efficiency of the different machines used in the experiment.

Conclusion:

Review Questions:1. If the ideal or theoretical mechanical advantage is increased, what is sacrificed? Do you lose speed?2. How is the theoretical mechanical advantage related with the number of strands supporting the weight?3. A system of pulleys has 6 supporting strands. If the efficiency of the arrangement is 75%, what is the actual mechanical advantage?4. What is the difference between the actual mechanical advantage and the theoretical mechanical advantage of any machine? Which of them is greater?5. Sketch a combination of pulleys where the theoretical mechanical advantage is 7.


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Date Performed:

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Experiment No. 8COEFFICIENT OF RESTITUTION

Objective:To be able to determine the coefficient of restitution of two colliding bodies.

Apparatus:steel ball, metal block, and meter stick

Theory:

The linear momentum of a body is defined as the product of its mass and velocity. It is shown by the equation:p = mvwhere:m is the mass of the body and v is the velocity

From Newtons Law of Motion, the conservation of momentum can be derived. Consider two bodies of masses m1 and m2 with velocities v1 and v2, respectively, and approaching each other in the same straight line (see Fig. 1). They collide in a direct central collision and no outside force acting on either of them except by either body on the other during impact. Fig. 2 shows the system in the process of collision where F1 is the impulsive force exerted by m2 on m1 and at the same time F2 is the impulsive force exerted by m1 on m2. F1 and F2 are equal in magnitude but opposite in direction. Fig. 3 shows the two bodies immediately after collision, moving away from each other with velocities u1 and u2.m1m2V1V2m2m1

Figure 1 Figure 2m2m1u1u2

Figure 3

The momentum of the two bodies before collision is and their momentum after collision is .

Applying the Principle of Conservation of Momentum, we have

Rearranging the terms, we obtainequation 1

If the two particles are perfectly elastic, energy will be conserved. The kinetic energy before collision is equal to the kinetic energy after collision.

Rearranging the terms, we obtainequation 2

Dividing equation 2 by equation 1, we have

where:v1 - v2 = the relative velocity of approachu2 - u1 = the relative velocity of separation

The ratio of the relative velocity of separation to that of the relative velocity of approach is called the coefficient of restitution between the two colliding bodies.

The coefficient of restitution e is the degree to which a pair of colliding bodies approaches perfect elasticity; that ise = 0 for perfectly inelastic bodiese = 1 for perfectly elastic bodies

Momentum is a vector quantity, so that the required sign must be observed in the equation for conservation of momentum. For instance, the algebraic signs of the momenta in Fig. 1 and Fig. 3 would be positive (+) for m1v1 , negative () for m2v2 , negative () for m1u1 , and positive (+) for m2u2 .

If the velocity is unknown and it comes out positive in the result of the solution., the assumed direction is correct; if negative, the direction of the velocity is opposite to that arbitrarily assumed. The magnitude of the velocity in each case is the correct value.

Procedures:1. Place a meter stick vertically on top of a solid metal block and drop a steel ball from a certain height. Practice doing this procedure and try to determine the height of rebound of the steel ball.2. When you have developed the skill in determining the height of rebound, begin with an initial height of fall at the 80-cm mark of the meter stick. Do this five times by decreasing the height of the fall by 10 cm and determine again the corresponding height of rebound.3. Compute the coefficient of restitution between the steel ball and the metal block using the equation .Since the metal block is stationary, then v2 and u2 are 0., v1=, and . Hence, we have

where:y1 is the height of fall of the steel ball, andy2 is the height of rebound of the steel ball

4. From the data obtained in step 3, plot the height of the fall along the abscissa scale (x-axis) and the height of rebound along the ordinate scale (y-axis).5. Measure the angle of the graph and determine the square root of the tangent of the angle that represents the coefficient of restitution between the two colliding bodies.

Data and Observations

TrialHeight of FallY1 (cm)Height of ReboundY2 (cm)Coefficient ofRestitution ( e )

180

270

360

450

540


Conclusion:

Review Questions:1. What are some of the probable reasons for the difference between the value of momentum before and after collision?2. The ballistic pendulum is employed to measure the velocity of a bullet. Explain its principle using the principle of conservation of momentum.3. Is momentum a vector or a scalar quantity?4. State the Law of Conservation of Momentum.5. On what physical principle or law is based the statement that the horizontal component of the velocity of a projectile remains constant?


Year & Section:

Date Performed:

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Experiment No. 9UNIFORM CIRCULAR MOTION

Objectives:1. To study the force involved in the motion of a body traveling with uniform speed along a circular path.

2. To determine the different factors affecting uniform circular motion.

Apparatus:rubber stopper with holes, hooked masses, meter stick, glass tube/ball pen tube set of masses, string and stopwatch

Theory:A body moving with constant speed along a circular path is being accelerated since the direction of its velocity changes. According to Newtons Second Law of Motion, there is a corresponding force that causes this acceleration, which is also constant in magnitude and is always directed towards the center. This force is known as central or centripetal force.

The set up consists of a glass tube with smooth edges. A string with one end tied to a rotating object is passed through the tube. The washers tied or suspended at the other end of a string serves as a tension when the body is rotating in horizontal circles.

Procedure:1. Record the mass of the rubber stopper.2. Tie a hooked mass of 200 g on one end of the string and pass it over the hole of the tube. Hold the tube vertically with the rubber stopper on top and hooked mass below.3. Supporting the 200 g mass with one hand and holding the glass tube in the other, hold the tube vertically and whirl the stopper in a horizontal circular motion, adjust the speed of rotation of the rubber stopper so that the hooked mass is just supported by the string.4. When the motion is under control, so that the hooked mass is stationary, one observer should count the number N of revolutions in one minute while the other maintains the constant motion. With the string in this position, measure the distance r from the top of the tube to the center of the rubber stopper.5. Using a constant hooked mass, vary the radius of this rotating object.6. Repeat step 2 using the hooked mass of 300 g and consider a constant radius of the string.


A. For 200 g.TrialsNo. of rev.NObserved Time t (s)Radius r (cm)VelocityV (cm/s)

1

2

3

Average

.

B. For 300 g.TrialsNo. of rev.NObserved Time t (s)Radius r (cm)VelocityV (cm/s)

1

2

3

Average

.


1. Compute the speed (V) of the stopper using

2. Compute the centripetal force using where:r = radius of rotationV = velocity of the stopperm = mass of the stopper

3. Determine the centripetal acceleration applying:

Conclusion:

Review Questions:1. When is an object said to be in uniform circular motion?2. Is there any unbalanced force acting on the body moving in a uniform circular motion? Explain.3. Could horizontal axis of rotation be used satisfactorily in an experiment on uniform circular motion? Explain.4. Upon what principle does the centrifuge work? Give examples of devices employing this principle.5. A person clings to a rotating merry-go-round to keep from being thrown from it. What kind of force do his muscles exert on his body?


Year & Section:

Date Performed:

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Experiment No. 10HOOKES LAW

Objective:To determine the relationship between the elongation of a coiled spring and stretching.

Apparatus:Hookes Law apparatus, set of weights and weight pan.

Theory:

Hookes Law apparatus is made up of coiled elastic spring which elongates when a pull of mass m is applied to its hooked end, where the pointer moves over a scale. Thus, elongation is indicated.

The force applied on the hook is called the stress, while the elongation of the spring is known as the strain. Stress is directly proportional to the strain if the spring returns to its original length upon the removal of the applied force.

Procedures:1. Hang the coiled elastic spring vertically. Attach the weight pan, if necessary. Then, place enough mass on it.2. Record the position of the weight pan with respect to the scale.3. Add 5 g at a time and record its new position each time you add.4. After adding all the required masses, remove the masses one at a time and record the resulting position.


Initial reading of the pointer = cm.

TrialsLoad(g)Reading of Pointer(cm)Displacement(cm)

LoadingUnloadingLoadingUnloading

1

2

3

4

5

Computation and Analysis:1. Construct a graph of displacement against the load during loading and unloading. Will one curve fit both sets of points?2. Describe the curve obtained and determine the relationship between the displacement and the stretching force.

Conclusion:

Review Questions:1. Why does additional stretching of a rubber band becomes more difficult at a greater rate after being extended or stretched to a certain length?2. After removing all the load from the weighing pan, does the spring return to its original length? If not, what could be the cause?3. Define elastic limit. How can the graph in your computation and analysis indicate that the elastic limit has been reached?

4. How does the force per unit obtained from the graph compare with the force constant of the spring?5. Does the graph indicate that the stress is directly proportional to the strain? Explain.

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