Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 07

We then define the work done by the constant force displace-as follows: Work done by a constant force: Is defined as the product of the component of the force in the direction of the dis

Fig 5.45 See Exercise (37)

mg

FD

t

y

(39) A canonical pendulum consists of a bob of mass m attached to the end of a

cord of length The bob whirls around in a horizontal circle of radius r at a

constant speedv while the cord always makes an angle θ with the vertical, see

Fig.5.46 Show that the bob’s speedv and period T (the time for one complete

revolution) are given by:

v = rg tan θ = g sin θ tan θ,

r

(40) What condition must be imposed on the relationship that governs the period of

a canonical pendulum in order to reach to the period of a simple pendulum?

Work, Energy, and Power 6

Work, energy, and power are words that have different meanings in our everydaylife Nevertheless, physicists give them specific definitions, which we present in thischapter

The work-energy power approach provides identical results to those obtained by

Newtonian mechanics, but usually with simpler analysis, especially when dealingwith complex situations where forces are not constant Therefore, we will introduce

two extremely important concepts: the work-energy-theorem and conservation of energy.

6.1 Work Done by a Constant Force

Consider a body that experiences a constant force →

F while undergoing a ment→s as it moves, see Fig.6.1 We then define the work done by the constant force

displace-as follows:

Work done by a constant force:

Is defined as the product of the component of the force in the direction of the

displacement and the magnitude of the displacement

Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_6,

© Springer-Verlag Berlin Heidelberg 2013

Fig 6.1 The work done by a

constant force →F while

The unit of work in SI units is N.m [abbreviated by joule (J)], i.e., 1 J = 1 N.m,

and in cgs units is dyne.cm (abbreviated by erg), i.e 1 erg= 1 dyne.cm Note that

1 J= 107erg, see Table6.1

Table 6.1 Units of work

Work Done by a Weight

Consider a block of mass m to be lifted up with almost zero acceleration (i.e., a  0)

Fig 6.2 Lifting a block with

almost zero acceleration

If the upward displacement of the block is denoted by→s , as in Fig.6.2, then we cancalculate the work done by →

Fas follows:

W F=F→•→s = Fs cos 0◦= Fs = m g s (6.3)where we have used the fact that the angle between the two parallel vectors →

Fand

→s is zero.

Also, we can calculate the work done by the gravitational force m→g as follows:

W g = m→g •→s = m g s cos 180◦= −m g s (6.4)Thus, we conclude that:

W F = m g s and W g = −m g s (Lifting case) (6.5)

where we have used the fact that the angle between the two antiparallel vectors m→gand→s is 180◦ The net work W

F +W gdone on the block is zero, as expected, becausethe net force on the block is zero This is not, of course, to say that it takes no work

to lift a block through a vertical height s In such a context, we do not refer to the net

work, but to the work done by the person

When we lower the block vertically downward with almost zero acceleration for

a displacement →s , see Fig.6.3, the sign of the work done by →

F and m→g will bereversed, since the sign of →s has reversed.

Following similar steps, one can easily find:

W F =F→•→s = F s cos 180◦= −F s = −m g s (6.6)

W g = m→g •→s = m g s cos 0◦= m g s (6.7)Thus, we conclude that:

W F = −m g s and W g = m g s (Lowering case) (6.8)

Fig 6.3 Lowering down a

block with almost zero

Final

Work Done by Friction

A common example in which the work is always negative is the work done byfriction When a block slides over a rough surface due to an applied force →

F , as

shown in Fig.6.4, the work done by the frictional force→

fkwhile the block undergoes

Fig 6.4 The work done by the kinetic frictional force →f

k while the block undergoes a displacement →s

is always negative and equals W f = −fks

From Fig.6.4, one can easily find the work done by gravity, the normal force, andthe applied force as follows:

fk, the force of gravity m→g , and the normal force N→ (b) Calculate the work done

of part (a) for m = 2 kg, μk = 0.5, θ = 30◦, F = 20 N, and d = 5 m.

Fig 6.5

Solution: (a) Since→

F is in the same direction as the displacement→

where h = yf − yi = d sin θ is the value of the vertical height That is, the work

done by gravity is negative and has a magnitude m g multiplied by height h This

result and Eq.6.4proves that the work is independent of the path taken betweenany two points

Since the force of friction →

fkis opposite to the displacement→

d , fk = μk N, and N = mg cos θ, the work done by friction will be:

6.2 Work Done by a Variable Force

One-Dimensional Analysis

Consider an object that is being displaced along the x-axis from xi to xf due to

the application of a varying positive force F(x), as shown in Fig.6.6a To calculatethe work done by this force, we imagine that the object undergoes a very smalldisplacementx from x to x + x due to the effect of an approximate constant force

F(x) as shown in Fig.6.6b For this very small displacement, we represent the amount

of work done by the force by the expression:

which is just the area of the magnified rectangle shown in Fig.6.6b Then, the total

work done from xito xfby the variable force F(x) is approximately equal to the sum

of the large number of rectangles in Fig.6.6b, i.e the total area under the force curve.Thus:

In the limit where x approaches zero, the value of the sum in the last

equa-tion approaches the exact value of the area under the force curve, see Fig.6.6c Asyou probably know from calculus, the limit of that sum is called an integral and isrepresented by:

Fig 6.6 (a) A variable force F(x) displaces a body in the positive x direction from xi to xf (b) The area

under the curve is divided into narrow strips of thicknessx, so that the approximate work done by the force F(x) for the small displacement x is W = F(x) x (c) In the limiting case, the work done by

the force is the colored area under the force curve

Therefore, we can express the work done by a variable force F(x) on an object that undergoes a displacement from xito xf as follows:

A force acting on an object varies with x as shown in Fig.6.7 Find the work done

by the force when the object undergoes a displacement from x = 0 to x = 7 m.

Fig 6.7

0 2 4 6

-2 -4

b

a

c H

Solution: The work done by the force equals the net signed area between the curve

and the displacement from x = 0 to x = 7 m That is, the area of the trapezoid minus

the area of the triangle Thus:

W = Area of the trapeziod − Area of the triangle

Work Done by a Spring

A spring is one type of common physical system in which the force (known asthe spring force) varies with position Figure6.8a, shows a massless block on a

horizontal frictionless surface attached to the free end of a relaxed spring If the

spring is stretched or compressed a small distance from equilibrium, the spring will

exert a force on the block This force is given by Hooke’s law as follows:

where x is the displacement of the block from its equilibrium position (x = 0) and

kH is a positive constant known as the spring constant (or the force constant).

The negative sign in Hooke’s law indicates that the direction of the force is always

opposite to the displacement The spring force is positive (to the right) when x < 0,

as in Fig.6.8b, and is negative (to the left) when x > 0, as in Fig.6.8c This type of

force always acts toward the equilibrium and is called a restoring force.

Fig 6.8 The variation of the

force of a spring on a block.

(a) When x= 0, the force is

zero (equilibrium position).

(b) When x is negative, the

force is positive (compressed

spring) (c) When x is positive,

the force is negative (stretched

spring) (d) Graph of F versus

x The work done by the spring

force as the block moves from

F x

F

xm Area

xx

Equilibrium position

-x

2

H m 1 2Area x

Frictionless

If we allow the block to compress the spring a distance of xmfrom its equilibriumposition and then release the block, it will move from−xmthrough the equilibrium

position x = 0 to xm In the absence of friction, the block will oscillate indefinitelybetween−xm and xm In this case xmis called the amplitude of the oscillations.

To calculate the work done by the spring force on the body as it moves from

xi= −xm to xf= 0, we use Hooke’s law in Eq.6.16as follows:

Note that the work done by the spring force is positive because the spring force

is in the same direction as the displacement We can reach the same result of Eq

6.18if we plot F versus x, as shown in Fig.6.8d, and then calculate the area of the

colored triangle that has a base xmand height kHxm On the other hand, when xi= 0

and xf= xm , we can find that Ws= −1

2kHx2m In this part of the motion, the springforce is to the left and the displacement is to the right, resulting in a negative work

Generally, if the block undergoes an arbitrary displacement from xito xf, the workdone by the spring force will be given by:

This shows that the work done is zero for any motion that has xi= xf

Let us calculate the work done by the applied force→

Fapp when the block moves

very slowly from xito xf, see Fig.6.9a and b To find this work we notice that→

Fapp

is equal and opposite to the spring force→

F at any displacement, i.e Fapp = −F =

Comparing Eq.6.19and Eq.6.20we find that W Fapp= − Ws, as expected If we plot

Fapp versus x, as shown in Fig.6.9c, then the work done by F in compressing the spring very slowly from xi= 0 to xf = −xmequals the area of the colored triangle

that has a base xmand heightkHxm, i.e Wapp=1

2kHx2m(Fappand the displacement are negative ).

Fig 6.9 (a) When x= 0, the

applied force is zero

(equilibrium position).

(b) When x is negative, the

applied force is negative

(compressed spring) (c) Graph

of the applied force versus x.

The work done by the applied

force as the block moves very

Frictionless

Example 6.3

An applied force Fapp= −5 N is exerted on the block that is attached to the free

end of the spring of Fig.6.9b As a result of this force, the spring is compressed

by 1 cm from its relaxed length (a) What is the spring constant of the spring? (b)What force does the spring exert on the block if the spring is compressed by 2.5cm? (c) How much work does the spring force do on the block as the spring iscompressed from the relaxed state by 2.5 cm? (d) How much work does the springforce do on the block during a total displacement starting from a compression of2.5 cm, passing through the equilibrium, and then to a stretch of 2.0 cm?

Solution: (a) The compressed spring pushes the block with a force F = −Fapp=

+5 N From F = −kH x, with x= −1 cm, we have:

(c) Since the spring is initially at its relaxed state, the work done by the spring

force on the block from xi= 0 to xf = −2.5 × 10−2m will be:

Ws =1

2kHx2i −12kHx2f = 0 −12(500 N/m)(−2.5 × 10−2m)2= −0.156 J

The work is negative because the spring force and the displacement are in oppositedirections Note that the amount of work done by the spring on the block would

be the same when stretching by 2.5 cm

(d) For this case, we have xi = −2.5 × 10−2m (the spring is initially

com-pressed) and xf = +2.0 × 10−2m (the spring is finally stretched) Then Eq.6.19becomes:

The work W done by the force →

F on the particle when it moves from an initial

position riof coordinates(xi, yi, zi) to a final position rf of coordinates(xf, yf, zf)

can be represented by:

Consider a particle of mass m, moving with acceleration a = a(x) along the x-axis

under the effect of a net force F(x) that points along this axis Thus, according to Newton’s second law of motion we have F (x) = ma The work done by this net force on the particle as it moves from an initial position xito a final position xfcan

be found using Eq.6.16as follows:

Note that when we change the integration variable from x to v we are required to

change the limits of the integration to the new variable

For a particle of mass m that has a speed v well below the speed of light, we define

its kinetic energy as:

Kinetic Energy

The kinetic energy K of a particle is defined as the product of one half of its

mass and the square of its speed, i.e

We can view kinetic energy as the energy associated with the motion of an object It

is more convenient to express Eq.6.29as:

where Kiis the particle’s initial kinetic energy and Kfis its final kinetic energy after

the work is done That is, the work done by the net force in displacing a particle

equals the change in its kinetic energy If there are many forces, such as an appliedforce→

F , a gravitational force m→g , a spring force F→s, a frictional force f→, etc, the work

done by the net force in displacing a particle will be equal to the sum of the workdone by all the forces acting on the particle That is:

Wnet = WF + Wg + Ws + Wf + = Kf − Ki = K (6.33)Equation6.32is known as the work-energy theorem This theorem is valid even

when the force varies in direction and magnitude while the particle (or the object)moves along an arbitrary curved path in three dimensions

Example 6.4

A box of mass m= 10 kg is initially at rest on a rough horizontal surface,

where the coefficient of kinetic friction between the box and the surface is

μk= 0.2 The box is then pulled horizontally by a force F = 50 N that makes

an angle θ = 60◦ with the horizontal, see Fig.6.10 (a) Use the work-energytheorem to find the speed vf of the box after it moves a distance s of 4 m

∗(b) Repeat part (a) using Newtonian mechanics.

Solution: (a) Both the weight-gravitational force m→g and the normal forceN→do

no work, since the displacement is horizontal, i.e W g = W N = 0 The work done

by the applied force is:

W F =F→•→s = F s cos θ = (50 N)(4 m)(cos 60◦) = 100 J

The magnitude of the frictional force is fk= μk N, where in this case

N = mg − F sin θ Therefore, the work done by friction is:

Thus, the acceleration of the box will be given by:

6.4 Conservative Forces and Potential Energy

In the previous section we introduced the concept of kinetic energy and found that itcan change only if work is done on the object In this section we introduce another

form of energy, called potential energy, associated with the position or configuration

of an object, and can be thought of as a stored energy that can be converted to kineticenergy or to work We begin by defining the following:

(a) Conservative and Non-conservative Forces

Conservative Forces

In Example 6.1 we were able to see that the work done by gravity depends only onthe initial and final vertical coordinates and hence is independent of the path takenbetween any two points Also, we found the same holds true in the case of a spring

In addition, we can easily see from Sect 6.2that the net work done on the object

by the gravitational force during a round trip is zero When a force exhibits these

properties, it is called a conservative force.

With reference to the arbitrary paths of Fig.6.11a, we can write the first conditionfor a conservative force as:

i.e., the work done by a conservative force on a particle moving from a to b along path 1 is the same as from a to b along path 2 In words:

That is, the work done by a conservative force on a particle that moves in a round trip

from a to b along path 1 and then from b to a along path 2 is zero In other words:

Soptlight

The net work done by a conservative force on a particle that is moving aroundany closed path is zero

From the work-energy theorem, W= 0 for a round trip, which means that the particle

will return to its starting point with the same kinetic energy it had when it started itsmotion