CHAPTER 7) WORK AND KINETIC ENERGY

7.1) Work Done by a Constant Force

• What is work? Consider Figure (7.1)

• A force is applied to an object, and the object slides along the horizontal direction.

• If we are interested in how effective the force is in moving the object, we consider (i) the magnitude of the force, and (ii) its direction.

• Assume that the magnitude of the applied force is the same in all three cases.

• The push applied in Figure (7.1b) does more to move the object than the push in Figure (7.1a).

• Figure (7.1c) – the applied force does not move the eraser at all, regardless of how hard it is pushed.

(a) (b) (c)

F FF

Figure 7.1

• In analyzing forces to determine the work required (to cause that motion) – consider (i) the vector nature of forces, and (ii) how far the object moves along the horizontal direction.

• Example – moving the eraser 3 m requires more work than moving it 2 cm.

• Figure (7.2) – an object undergoes a displacement d along a straight line while acted on by a constant force F that makes an angle with d.

• The work W done on an object by an agent exerting a constant force on the object is the product of the component of the force in the direction of the displacement and the magnitude of the displacement:

The situation where a force does no work (W=0)

1. If the object does not move – i.e if d=0, Equation (7.1) gives W = 0 (the situation in Figure (7.1c)). Example – holding a chair.

2. When the force applied is perpendicular to the object’s displacement – i.e. if = 90o, then from Eq. (7.1), W = 0 because cos 90o = 0. Example: Figure (7.3) – the work done by the normal force on the object and the work done by the force of gravity on the object = zero because both forces are perpendicular to the displacement and have zero components in the direction of d.

cosFdW (7.1)

The sign of the work done by the applied force

• Depends on the direction of F relative to d.

• Positive when the vector associated with the component F cos is in the same direction as the displacement.

Example – when an object is lifted, the work done by the applied force is positive because the direction of that force is upward, the same direction as the displacement.

• Negative when the vector associated with the component F cos is in the direction opposite the displacement.

Example – as an object is lifted, the work done by the gravitational force on the object is negative.

• The factor cos in the definition of W (Eq. (7.1)) automatically takes care of the sign.

• Work is an energy transfer; if energy is transferred to the system (object), W is positive; if energy is transferred from the system, W is negative.

• If an applied force F acts along the direction of the displacement, then = 0 and cos 0o = 1.

• In this case, Equation (7.1) gives : W = Fd

• Work is a scalar quantity, and its units are force multiplied by length.

• The SI unit of work = newton·meter (N·m) = joule (J).

• In general, a particle may be moving with either a constant or a varying velocity under the influence of several forces.

• In these cases, because work is a scalar quantity, the total work done as the particle undergoes some displacement is the algebraic sum of the amounts of work done by all the forces.

Example (7.1) : Mr. Clean

A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F = 50.0 N at an angle of 30.0o with the horizontal (Fig. (7.4a)). Calculate the work done by the force on the vacuum cleaner as the vacuum cleaner is displaced 3.00 m to the right.

7.2) The Scalar Product of Two Vectors

Scalar Product

• Indicate how F and d interact in a way that depends on how close to parallel they happen to be.

• Scalar product = F·d (Also known as the dot product – because of the dot symbol).

• Equation (7.1) becomes : W = F·d = Fd cos (7.2)

• F·d (read “F dot d”) is a shorthand notation for Fd cos .

• In general, the scalar product of any two vectros A and B is a scalar quantity equal to the product of the magnitudes of the two vectors and the cosine of the angle between them : A ·B AB cos (7.3)

• A and B need not have the same units.

• Figure (7.6) – B cos is the projection of B onto A. Equation (7.3) sayas that A ·B is the product of the magnitude of A and the projection of B onto A.

• The scalar product is commutative : A ·B = B ·A

That is, the order of the dot product can be reversed.

• The scalar product obeys the distributive law of multiplication :

A ·( B + C ) = A ·B + A ·C

• If A is perpendicular to B ( = 90o), then A ·B = 0.

• The case of A ·B =0, when either A or B is zero.

• If vector A is parallel to vector B and the two point in the same direction ( = 0o) , then A ·B = AB.

• If vector A is parallel to vector B but the two point in opposite directions ( = 180o), then A ·B = - AB.

• The scalar product is negative when 90o < < 180o.

• The unit vector i, j, and k – lie in the positiove x, y, and z directions, respectively, of a right-handed coordinate system.

• It follows from the definition of A ·B that the scalar products of these unit vectors are : i ·i = j ·j = k ·k = 1 (7.4)

i ·j = i ·k = j ·k = 0 (7.5)

• Two vectors A and B can be expressed in component vector form as :

A = Ax i + Ay j + Az k

B = Bx i + By j + Bz k

• Using the information given in Equations (7.4) and (7.5) – the scalar product of A and B reduces to :

A ·B = Ax Bx + Ay By + Az Bz (7.6)

• In special case in which A = B : 22

z2y

2x AAAA AA

Example (7.2) : The Scalar Product

The vector A and B are given by A =2i + 3j and B =-i + 2j. (a) Determine the scalar product A ·B.

Example (7.3) : Work Done by a Constant Force

A particle moving in the xy plane undergoes a displacement d = (2.0i + 3.0j) m as a constant force F = (5.0i + 2.0j) N acts on the particle. (a) Calculate the magnitude of the displacement and that of the force.

7.3) Work Done by a Varying Force

• Consider a particle being displaced along the x axis under the action of a varying force.

• The particle is displaced in the direction of increasing x from x = xi to x = xf.

• In such a situation, we cannot use W = (F cos ) d to calculate the work done by the force because this relationship applies only when F is constant in magnitude and direction.

• If we imagine that the particle undergoes a very small displacement x (Figure (7.7a)) – the x component of the force Fx is approximately constant over this interval.

• For this small displacement, we can express the work done by the force as :

• Imagine that the Fx versus x curve is divided into a large number of such intervals.

• Then the total work done for the displacement from xi to xf is approximatelay equal to the sum of a large number of such terms :

xFW x The area of the shaded rectangle in Fig. 7.7a

f

i

x

xx xFW

• If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit but the value of the sum approaches a definite value equal to the area bounded by the Fx curve and the x axis :

• This definite integral is numerically equal to the area under the Fx-versus-x curve between xi and xf.

• Therefore, we can express the work done by Fx as the particle moves from xi to xf as :

• If more than one force acts on a particle, the total work done is just the work done by the resultant force.

• If we express the resultant force in the x direction as Fx , then the total work, or net work, done as the particle moves from xi to xf is :

f

i

f

i

x

xx

x

xx dxFxFlim

x 0

f

i

x

xxdxFW (7.7) Work done by a varying force

f

i

x

xxnet dxFWW (7.8)

Example (7.4) : Calculating Total Work Done from a Graph

A force acting on a particle varies with x, as shown in Figure (7.8). Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m.

Example : A block of mass 2.5 kg is pushed 2.2 m along a frictionless horizontal table by a constant 16.0-N force directed 25o below the horizontal. Determine the work done by (a) the applied force, (b) the normal force exerted by the table, (c) the force of gravity, and (d) the net force on the block.

Fx(N)

5

01 2 3 4 5 6

x(m)

A B

C

Figure (7.8)

Example : A force F = (6i – 2j) N acts on a particle that undergoes a displacement s = (3i + j) m. Find (a) the work done by the force on the particle and (b) the angle between F and s.

Example : The force acting on a particle varies as in Figure (P7.24). Find the work done by the force as the particle moves (a) from x = 0 to x = 8.0 m, (b) from x = 8.0 m to x = 10 m, and (c) from x = 0 to x = 10 m.

2 4 6 8 10 12x (m)0

6

-2

A

B

C

D

E

Fx (N)

Work Done by a Spring

• Figure (7.10) – A common physical system for which the force varies with position.

• A block on a horizontal, frictionless surface is connected to a spring.

• If the spring is either stretched of compressed a small distance from its unstretched (equilibrium) configuration, it exerts on the block a force of magnitude :

where x = the displacement of the block from its unstretched (x=0) position

k = a positive constant called the force constant of the spring.

• The force required to stretch or compress a spring is proportional to the amount of stretch or compression x.

• This force law for springs = Hooke’s law (valid only in the limiting case of small displacements).

• The value of k is a measure of the stiffness of the spring. Stiff springs have large k values, and soft springs have small k values.

kxFs (7.9) Spring force

• The negative sign (Eq. (7.9)) – signifies that the force exerted by the spring is always directed opposite the displacement.

• When x > 0 (Figure (7.10a)) – the spring force is directed to the left, in the negative x direction.

• When x < 0 (Figure (7.10c)) – the spring force is directed to the right, in the positive x direction.

• When x = 0 (Figure (7.10c)) – the spring is unstretched and Fs = 0.

• Because the spring force always acts toward the equilibrium position (x = 0), it sometimes is called a restoring force.

• If the spring is compressed until the block is at the point -xmax and then released the block moves from -xmax through zero to +xmax.

• If the spring is stretched until the block is at the point xmax and is then released the block moves from +xmax through zero to -xmax.

• It then reverses direction, returns to +xmax , and continues oscillating back and forth.

Suppose the block has been pushed to the left a distance xmax from equilibrium and is then released

• Calculate the work Ws done by the spring force as the block moves from xi = -xmax to xf = 0.

• Apply Equation (7.7), and assume the block may be treated as a particle :

• The work done by the spring force is positive because the force is in the same direction as the displacement (both are to the right).

• Consider the work done by the spring force as the block moves from xi = 0 to xf = xmax.

• We find that because the displacement is to the right and the spring force is to the left.

• Therefore, the net work done by the spring force as the block moves from xi = - xmax to xf = xmax is zero.

f

i max

x

x

0

x

2max2

1ss kxdx)kx(dxFW (7.10)

2max2

1s kxW

• Figure (7.10d) – a plot of Fs versus x.

• The work calculate in Eq. (7.10) is the area of the shaded triangle, corresponding to the displacement from -xmax to 0.

• Because the triangle has base xmax and height kxmax, its area is ½ kx2max , the

work done by the spring as given by Eq. (7.10).

• If the block undergoes an arbitrary displacement from x=xi to x=xf , the work done by the spring force is :

• From Eq. (7.11), the work done by the spring force is zero (Ws = 0) – for any motion that ends where it began (xi = xf).

2f2

1x

x

2i2

1s kxkxdx)kx(W

f

i

(7.11) Work done by a spring

Work done on the spring by an external agent

• The external agent stretches the spring very slowly form xi = 0 to xf = xmax (Figure (7.11)).

• Calculate the work by noting that at any value of the displacement, the applied force Fapp is equal to and opposite the spring force Fs , so that Fapp = - (-kx) = kx.

• The work done by this applied force (the ecternal agent) is :

2max2

1x

0

x

0appF kxkxdxdxFW

maxmax

app

This work is equal to the negative of the work done by the spring force for this displacement

Figure (7.11)

FappFs

xf = xmaxxi = 0

Example (7.6) : Measuring k for a Spring

• A common technique used to measure the force constant of a spring is described in Figure (7.12).

• The spring is hung vertically, and an object of mass m is attached to its lower end.

• Under the action of the “load” mg, the spring stretches a distance d from its equilibrium position.

• Because the spring force is upward (opposite the displacement) it must balance the downward force of gravity mg when the system is at rest.

• In this case, we apply Hooke’s law to give |Fs| = kd = mg , or

• For example, if a spring is stretched 2.0 cm by a suspended object having a mass of 0.55 kg, then the force constant is :

d

mgk

m/N107.2

m100.2

)s/m80.9)(kg55.0(

d

mgk

2

2

2

d

Fs

mg

(a)

(b)

(c)Figure (7.12)

7.4) Kinetic Energy andThe Work-Kinetic Energy Theorem

• Figure (7.13) - a particle of mass m moving to the right under the action of a constant net force F.

• Because the force is constant - from Newton’s second law, the particle moves with a constant acceleration a.

• If the particle is displaced a distance d, the net work done by the total force F is :

• The relationships below are valid when a particle undergoes constant acceleration :

where vi = the speed at t = 0 and vf = the speed at time t.

d)ma(dFW (7.12)

t

vva

t)vv(d

if

fi21

m

d

F

vi vf

Figure (7.13)

• Substituting these expressions into Equation (7.12) :

• The quantity represents the energy associated with the motion of the particle = kinetic energy.

• The net work done on a particle by a constant net force F acting on it equals the change in kinetic energy of the particle.

• In general, the kinetic energy K of a particle of mass m moving with a speed v is defined as :

• Kinetic energy is a scalar quantity and has the same units as work = Joule (J).

• It is often convenient to write Equation (7.13) in the form :

2i2

12f2

1

fi21if

mvmvW

tvvt

vvmW

(7.13)

221 mv

221 mvK (7.14)

KKKW if (7.15) Work-kinetic energy theorem

fi KWK • That is :

Work-kinetic energy theorem

• Must include all of the forces that do work on the particle in the calculation of the net work done.

• The speed of a particle increases if the net work done on it is positive - because the final kinetic energy is greater than the initial kinetic energy.

• The speed of a particle decreases if the net work done is negative because the final kinetic energy is less than the initial kinetic energy.

• Kinetic energy is the work a particle can do in coming to rest, or the amount of energy stored in the particle.

• We derived the wrok-kinetic energy theorem under the assumption of a constant net force - but it also is valid when the force varies.

• Suppose the net force acting on a particle in the x direction is Fx.

• Apply Newton’s second law, Fx = max, and use Equation (7.8), the net work done :

f

i

f

i

x

x

x

x

x

x dxmadxFW

• If the resultant force varies with x, the acceleration and speeed also depend on x.

• Acceleration is expressed as :

• Substituting this expression for a into the above equation for W gives :

• Situation Involving Kinetic Friction

• Analyze the motion of an object sliding on a horizontal surface - to describe the kinetic energy lost because of friction.

• Suppose a book moving on a horizontal surface is given an initial horizontal velocity vi and slides a distance d before reaching a final velocity vf (Figure (7.15)).

dx

dvv

dt

dx

dx

dv

dt

dva

2i2

12f2

1

v

v

x

x

mvmvW

mvdvdxdx

dvmvW

f

i

f

i

(7.16)

The net work done on a particle by the net force acting on it is equal to the change in its kinetic energy of the particle

fk

vi vf

d Figure (7.15)

• The external force that causes the book to undergo an acceleration in the negative x direction is the force of kinetic friction fk acting to the left, opposite the motion.

• The initial kinetic energy of the book is ½mvi2 , and its final kinetic energy is

½mvf2 (Apply the Newton’s second law).

• Because the only force acting on the book in the x direction is the friction force, Newton’s second law gives -fk = max.

• Multiplying both sides of this expression by d and using Eq. (2.12) in the form vxf

2 - vxi2 = 2axd for motion under constant acceleration give

-fkd = (max)d = ½mvxf2 - ½mvxi

2 or : dfK kfriction (7.17a)

• The amount by which the force of kinetic friction changes the kinetic energy of the book is equal to -fkd.

• Part of this lost kinetic energy goes into warming up the book, and the rest goes into warming up the surface over which the book slides.

• The quantity -fkd is equal to the work done by kinetic friction on the book plus the work done by kinetic friction on the surface.

• When friction (as welll as other forces) - acts on an object, the work-kinetic energy theorem reads :

Wother represents the sum of the amounts of work done on the object by forces other than kinetic friction.

fkotheri KdfWK (7.17b)

Example (7.7) : A Block Pulled on a Frictionless Surface

A 6.0-kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3.0 m.

n vf

F

dmg

(a)

n vf

F

dmg

fk

(b)

Figure (7.16)

Example (7.8) : A Block Pulled on a Rough Surface

Find the final speed of the block described in Example (7.7) if the surface is not frictionless but instead has a coefficient of kinetic friction of 0.15.

Example (7.9) : Does the Ramp Lessen the Work Required?

A man wishes to load a refrigerator onto a truck using a ramp, as shown in Figure (7.17). He claims that less work would be required to load the truck if the length L of the ramp were increased. Is his statement valid?

Example (7.10) : Useful Physics for Safer Driving

A certain car traveling at an initial speed v slides a distance d to a halt after its brakes lock. Assuming that the car’s initial speed is instead 2v at the moment the brakes lock, estimate the distance it slides.

L

Figure (7.17)

Example (7.11) : A Block-Spring System

A block of mass 1.6 kg is attached to a horizontal spring that has a force constant of 1.0x103 N/m (Figure (7.10)). The spring is compressed 2.0 cm and is then released from rest. (a) Calculate the speed of the block as it passes through the equilibrium position x = 0 if the surface is frictionless.

7.5) Power

• The two cars - has different engines (8 cylinder powerplant and 4 cylinder engine), same mass.

• Both cars climb a roadway up a hill, but the car with the 8 cylinder engine takes much less time to reach the top.

• Both cars have done the same amount of work against gravity, but in different time periods.

• Know (i) the work done by the vehicles, and (ii) the rate at which it is done.

• The time rate of doing work = power.

• If an external force is applied to an object (assume acts as a particle), and if the work done by this force in the time interval t is W, then the average power expended during this interval is defined as :

• The work done on the object contributes to the increase in the energy of the object.

• Therefore, a more general definition of power is the time rate of energy transfer.

• We can define the instantaneous power P as the limiting value of the average power as t approaches zero :

t

W

Average power

dt

dW

t

Wlim

t 0Represent increment of work done by dW

• From Equation (7.2), letting the displacement be ecpressed as ds, that dW = F·ds.

• Therefore, the instantaneous power can be written :

• The SI unit of power = joules per second (J /s) = watt (W).

• A unit of power in the British engineering system is the horsepower (hp) :

• A unit of energy (or work) can now be defined in terms of the unit of power.

• One kilowatt hour (kWh) = the energy converted or consumed in 1 h at the constant rate of 1 kW = 1 000 J /s.

vFs

F dt

d

dt

dW(7.18) v =ds / dt

1W = 1 J/s = 1 kg ·m2 / s3

1 hp = 746 W

1kWh = (103W)(3600 s) = 3.60 x 106 J kWh = a unit of energy, not power

Example (7.12) : Power Delivered by an Elevator Motor

An elevator car has a mass of 1 000 kg and is carrying passengers having a combined mass of 800 kg. A constant frictional force of 4 000 N retards its motion upward, as shown in Figure (7.18a). (a) What must be the minimum power delivered by the motor to lift the elevator car at a constant speed of 3.00 m/s?