MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

Physics 8.01 Fall 2012

Problem Set 5 Work and Kinetic Energy Solutions

Problem 1: Work Done by Forces a) Two people push in opposite directions on a block that sits atop a frictionless surface (The soles of their shoes are glued to the frictionless surface). If the block, originally at rest at point P, moves to the right without rotating and ends up at rest at point Q, describe qualitatively how much work is done on the block by person 1 relative to that done by person 2?

Solution: Initially the block is at rest. After the pushing has ended, the block ends at rest, so the change in kinetic energy is zero. From the work-kinetic energy theorem, this implies that the total work done on the block is zero. The total work done on the block is the sum of the work done on the block by each person. Since the block moves to the right, person who pushes the block to the right does a positive work, and the person pushing to the left does negative work. Since the total is zero, the magnitude of the work done by each person is equal. b) Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the distances needed to stop them compare? Explain your reasoning. Answer 3. The kinetic energy of an object can be written as 2 / 2K p m= . Because the ping pong ball and the bowling ball have the same momentum, the kinetic energy of the less massive ping pong ball is greater than the kinetic energy of the more massive bowling ball. You must do work on an object to change its kinetic energy. If you exert a constant force, then the work done is the product of the force with the displacement of the point of application of the force. Since the work done on an object is equal to the change in kinetic energy, the ping pong ball has a greater change in kinetic energy in order to bring it to a stop, so the you need a longer distance to stop the ping pong ball.

Alternative Solution: Both the initial momentum and the force acting on the two objects are equal. Therefore the initial velocity and the acceleration of the ping-pong ball isgreater than the bowling ball by the ratio of the bowling ball mass to the ping-pong ball mass.

0, 0,p p b b xm v m v p= =

p p b b xm a m a F= = Since both the force acting on each object and the change in momentum is the same, the impulse acting on each ball is the same. Therefore, the time interval it takes to stop each object is the same. Since the displacement is equal to

,0 2x

xa t

x v t!" #! = $ !% &' (

The ratio p b

b p

x mx m

!=

!

hence the ping-pong has the greater displacement.

Problem 2: Kinetic Energy and Work

A tetherball of mass m is attached to a post of radius b by a string. Initially it is a distance r0 from the center of the post and it is moving tangentially with a speed v0 . Ignore gravity and any dissipative forces.

a) Suppose the string passes through a hole in the center of the post at the top and is gradually shortened by drawing it through the hole (figure above left). Until the ball hits the post, is the kinetic energy of the ball constant? Explain your reasoning.

Answer: Since the path of the ball is not circular, a small displacement of the ball has a radial component inward so the dot product between the force by the rope on the ball with the displacement is non-zero hence the work done by the force is not zero. Therefore the kinetic energy of the ball changes.

.

b) Now suppose that the string wraps around the outside of the post (figure above right). Until the ball hits the post, is the kinetic energy of the ball constant? Explain your reasoning.

Answer A small displacement of the ball is always perpendicular to string since at each instant in time the ball undergoes an instantaneous circular motion about the string contact point with pole. Therefore the dot product between the force by the rope on the ball with the displacement is zero hence the work done by the force is zero. Therefore the kinetic energy of the ball does not change.

Problem 3: An object of mass m = 4.0kg , starting from rest, slides down an inclined plane of length l = 3.0m . The plane is inclined by an angle of ! = 300 to the ground. The coefficient of kinetic friction µk = 0.2 . At the bottom of the plane, the mass slides along a rough surface with a coefficient of kinetic friction µk (x) = (0.05 m

!1)x until it comes to rest. The goal of this problem is to find out how far the object slides along the rough surface.

a) What is the work done by the friction force while the mass is sliding down the

inclined plane? Is this positive or negative?

b) What is the work done by the gravitational force while the mass is sliding down the inclined plane? Is this positive or negative?

c) What is the kinetic energy of the mass just at the bottom of the inclined plane?

d) What is the work done by the friction force while the mass is sliding along the

ground? Is this positive or negative?

e) How far does the object slide along the rough surface? Solution: a) While the object is sliding down the inclined plane the kinetic energy is increasing due to the positive work done on the object by the gravitational force and the negative work (smaller in magnitude) done by friction force. As the object slides along the level surface, the (negative) work done by the friction force slows the object down. We will use the work-kinetic energy theorem to calculate the change in kinetic energy for each stage. The free body diagram on the inclined plane and on the level surface are shown below.

Choose a coordinate system with the origin at the top of the inclined plane and the positive x-direction pointing down the inclined plane. Then the work done by the friction force is

0

0

friction0

( ) ( ) cos 0x l

x k inclined k inclinedx

W d F dx Nl mg lµ µ !=

=

= " = = # = # <$ $F r! !

Wfriction = !(0.2)(4.0kg)(9.8m " s-2 )(3.0m)(cos(30o ) = !20.4 J Note that the normal force is determined from Newton’s Second Law applied to the normal direction to the inclined plane.

N ! mg cos" = 0 b) The work done by the gravitational force is just

Wgrav = !mg(h f ! h0 )

note that (hf ! h0 ) = -l sin" . So the work done by the gravitation force is

Wgrav = mgl sin! > 0

The magnitude of this work is

Wgrav = (4.0kg)(9.8 m ! s-2 )(3.0m)(sin(30o ) = 58.8 J

c) The total work is

Wtotal =Wgrav +Wfriction = mgl(sin! " (µk )inclined cos!)Wtotal = 58.8 J " 20.4 J = 38.4 J

The change in kinetic energy is just equal to the final kinetic energy at the bottom of the incline because the started from rest,

!K =12mv2bottom

So the work-kinetic energy theorem Wtotal = !K becomes

mgl(sin! " (µk )inclined cos!) =12mv2bottom = 38.4 J

e) Choose a coordinate system with the origin at the base of the inclined plane and the positive x-direction pointing in the direction the object moves along the plane. The

normal force on the object is determined from Newton’s Second Law applied to the normal direction to the plane.

N ! mg = 0 Then the work done by the friction force is

Wfriction =!F ! d!r" = Fx dx

x0 =0

x0 =d

" = (0.05 m#1)Nx dxx0 =0

x0 =d

"

Wfriction = #(0.05 m#1)Nd 2

2= #(0.05 m#1)mg

d 2

2< 0

f) The change in kinetic energy is just equal to the kinetic energy at the bottom of the incline plane because the object comes to rest

!K = "12mv2bottom

So the work-kinetic energy theorem Wtotal = !K becomes

!

12

mv2bottom = !(0.05 m!1)mg

d 2

2

However we have already determined this kinetic energy so this last equation becomes

!mgl(sin" ! (µk )inclined cos") = !(0.05 m!1)mg d 2

2

we can now solve for the distance it traveled before it came to rest along the horizontal

d =

2(mgl(sin! " (µk )inclined cos!))(0.05 m"1)mg

=2(38.4 J)

(0.05 m"1)(4.0kg)(9.8m # s-2 )= 6.3 m

Note: If we consider the entire motion from the release at the top of the inclined plane to coming to rest on the horizontal, then the total change in kinetic energy is zero and

0 = !K =Wtotal =Wgrav + (Wfriction )inclined + (Wfriction )horizontal Thus

0 = mgl sin! " mgl(µk )inclined cos! " (0.05 m"1)mg d2

2

which leads to the identical expression for the distance traveled on the horizontal before coming to rest.

Problem 4: Asteroid about Sun

An asteroid of mass m is in a non-circular closed orbit about the sun. Initially it is a distance ir from the sun, with speed iv . What is the change in the kinetic energy of the asteroid when it is a distance is

rf , from the sun? Solution: We shall use the work-kinetic energy theorem,

2 21 12 2f i f iW K K K mv mv= ! " # = #

The work done by the gravitational force is the line integral

f

i

r

grW d= !" F r

! !.

Let’s choose polar coordinates.

The gravitational force between the sun and the asteroid is given by

sung 2

Gm m ˆr

= !F r!

.

As the asteroid moves, the infinitesimal displacement is tangent to the path and is given in polar coordinates by

ˆˆd dr rd!= +r r è! The work done by the gravitational force on the body is given by the line integral

sun2 ( )f f

i i

r r

gr r

Gm m ˆˆ ˆW d dr rdr

!= " = # " +$ $F r r r è! ! .

Note that because 0ˆˆ ! =r è and 1ˆ ˆ! =r r , only the radial part of the displacement contributes to the work done by the gravitational force,

sun2

f

i

r

r

Gm mW drr

= !" .

Upon evaluation of this integral, we have for the work

sun sunsun2

1 1ff

ii

rr

rr f i

Gm m Gm mW dr Gm mr r r r

! "= # = = #$ %$ %& '( .

Using the work-kinetic energy theorem, the change in kinetic energy is

sun1 1

f i

K Gm mr r

! "# = $% &% &' (

.

The final kinetic energy of the asteroid is then

2 2sun

1 1 1 12 2f i

f i

mv mv Gm mr r

! "= + + #$ %$ %& '

Let’s check our result by considering the case that the asteroid is moving closer to the sun, f ir r< , hence 1 1f i/ r / r> . Thus the work done by gravitational force on the asteroid is positive,

sun0

1 1 0f

W Gm mr r

! "= # >$ %$ %& '

.

and the kinetic energy of the asteroid increases as we expect for a body moving closer to the sun.

Problem 5 Work Done by a Several Forces A block of mass m slides along a horizontal table with speed v0 . At x = 0 it hits a spring with spring constant k and begins to experience a friction force. The coefficient of friction is given by µ . How far did the spring compress when the block first momentarily comes to rest.

Solution: From the model given for the frictional force, we could find the non-conservative work done, which is the same as the loss of mechanical energy, if we knew the position

x f

where the block first comes to rest. The most direct (and easiest) way to find x f is to use

the work-energy theorem,

2 20

0

1 12 2

fx x

x f i fx

F dx K K K mv mv=

=

= ! " # = #$

Since we are trying to find the distance that the object moved when it first becomes to rest, we have that 0fv = , so the work-kinetic energy theorem becomes

20

0

12

fx x

xx

F dx mv=

=

=! .

There are two forces acting on the block, friction and the spring force. The x-component of these forces are given by

x x ,spring x , fricitonF F F kx mgµ= + = ! ! . So the work done on the object is

x x,spring x,friction0 0 0

0 0

212

f f f

f f

x x x x x x

x x x

x x

f f

W F dx F dx F dx

kx dx mg dx

kx mgx

µ

µ

= = =

= = =

= = +

= ! !

= ! !

" " "

" " (1)

Applying the work energy theorem yields

2 20

1 12 2f fk x mgx mvµ! ! = ! (2)

We can rearrange this as 2 2

02 0f fmg m

x x vk k

µ+ ! = .

The solution of this quadratic equation is given by

x f = !

µmgk

±µmg

k"#$

%&'

2

+mk

v02

Note that we have assumed that 0fx > , therefore we need to choose the positive square root,

x f = !

µmgk

+µmg

k"#$

%&'

2

+mk

v02 .

It is worth checking that the above result is dimensionally correct. Recall that Hooke’s law states that F kx= ! , so /mg kµ has the dimensions of length. Similarly 2

0mv has

the dimensions of energy or force times distance. So mk

v02 has the dimensions of length

squared.

Problem 6 Sticky Pendulum A simple pendulum consists of a bob of mass m1 that is suspended from a pivot by a string of length l and negligible mass. The bob is pulled out and released from a height h0 as measured from the bob’s lowest point directly under the pivot point and then swings downward in a circular orbit as shown in the figure below. At the bottom of the swing, the bob collides with a block of mass m2 that is initially at rest on a frictionless table. Assume that there is no friction at the pivot point.

a) What is the work done by the gravitational force on the bob from the instant when the bob is released to the instant just before the collision?

b) How much work does the tension force do as the bob moves in a circular path? c) What is the speed of the bob at the bottom of the swing immediately before the

collision? d) Suppose the bob and block stick together after the collision. What is the speed of the

combined system immediately after the collision? e) What is the tension in the string immediately after the collision? f) What is the change in kinetic energy of the block and bob due to the collision? What

is the ratio of the change in kinetic energy to the kinetic energy before the collision? Solution:

a) Choose a coordinate system with the origin at the bottom of the swing as shown in the figure below. The gravitational force is m

!g = !mgj . The displacement is

d!r = dx i + dy j . Therefore the work done by the gravitational force is

W g = m!g ! d!r

i

f

" = #mgj ! (dx i + d $y j)i

f

" = #mg d $y$y = yi

$y = y f

" = #mg( y f # yi ) = mgh0

b) The tension force does zero work because it is perpendicular to the displacement of the

bob.

c) We can use the work energy theorem, W g = !K , noting that the bob started from rest

m1 g h0 =

12

m1 v1, b2 (3)

Therefore the speed of the bob at the low point of the swing just before the collision,

v1, b = 2 g h0 . (4)

d) Consider the bob and the block as the system. Although tension in the string and the gravitation force are now acting as external forces, both are particular to the motion of the bob and block during the collision. If we additionally assume that the collision is nearly instantaneous, then the momentum is constant in the direction of the bob’s motion,

m1 v1, b = (m1+ m2 )va , (5) where va is the speed of the bob and block immediately after the collision. Therefore

va =

m1

m1+ m2

v1, b . (6)

Using Eq. (4) in Eq. (6) yields

va =

m1

m1+ m2

2gh0 . (7)

d) We show the free body force diagram immediately after the collision. Newton’s

Second Law becomes

!T + 4m1g = !4m1

va2

l (8)

Therefore substitute Eq. (7) into Eq. (8) and solve for the tension in the string immediately after the collision

T = 4m1g 1+

2h0

lm1

2

(m1+ m2 )2

!

"#

$

%& (9)

e) The change in kinetic energy of the bob and block due to the collision in part c) is given by

!K = Kafter " Kbefore =

12

4m1va2 "

12

m1v1,b2 . (10)

Using Eq. (7), and (4) in Eq. (10) yields

!K = Kafter " Kbefore =12

4m1va2 "

12

m1v1,b2

!K =2m1m1

2 2gh0

(m1+ m2 )2"

12

m12gh0 = m1gh0

4m12

(m1+ m2 )2"1

#

$%

&

'(

(11)

The kinetic energy before the collision was 1 0m gh , and so the ratio of the change in kinetic energy to the kinetic energy before the collision is

!KKbefore

=4m1

2

(m1+ m2 )2"1

#

$%

&

'( . (12)