Download 1. A good, professional baseball pitcher throws a ball straight up in

1. A good, professional baseball pitcher throws a ball straight up in the air. Using the
Work-Kinetic Energy theorem, estimate how high the ball will go. (A good throw can
reach 90 mph.)
 Assume the ball is thrown with an initial velocity of v0 = 90mph = 40 m s .
 Upon release, the only force acting on the ball is the force of gravity, this force does
work on the ball as the ball moves up to its maximum height.
 W = !K
! !
 W = Fi!y = "mg!y  work done by gravity on the ball.
!mg"y = 12 m v 2f ! 12 mv02
!"
=0
!g"y = ! 12 v02
v02
( 40 m s ) = 82m
=
2g 2 ( 9.8 m s2 )
2
!y =
2. An object is moving in such a way that the force of gravity is increasing its kinetic
energy. You can conclude that
1. the object is falling freely
2. there is a net force on the object
3. both A and B
4. neither A nor B
 If the work done by gravity is causing an increase in Kinetic energy, then the force of
gravity and the displacement of the object are in the same direction: Wgrav > 0
 If the Kinetic energy is increasing, then the velocity is increasing. This means there
must be an acceleration, which only comes from a net force acting on the object.
3. Can static friction do work? If so, give an example. [Hint: Static friction acts to prevent
relative motion along the contact surface.]
4. You need to move a heavy crate by sliding it across the floor. The coefficient of
sliding (kinetic) friction is 0.2. You can either push the crate horizontally or pull the crate
using an attached rope. When you pull on the rope, it makes a 30° angle with the floor.
Which way should you choose to move the crate so that you will do the least amount of
work? How can you answer this question without knowing the weight of the crate or the
displacement of the crate?
Let’s assume you are pushing the crate at a constant
speed.
!
!
Fpush = fk = µ k N = µ k mg
!
!
W push = Fpush i!x = µ k mg!x
!F
y
= N + Fpull sin 30° " mg = 0
N = mg ! Fpull sin 30°
!F
x
= Fpull cos 30° " fk = 0
(
Fpull = µ k mg ! Fpull sin 30°
Fpull =
!
!
W pull = Fpull i!x =
)
µ k mg
(1 + µk sin 30°)
µ k mg
!x
(1 + µk sin 30°)
(1 + µk sin 30°) = 1.1

W pull =
W push
µ k mg
!x =
1.1
1.1
!
!
For Fpush = Fpull  you do less work on the crate when you pull it with a rope.
Work done by a spring:
Recall Hooke’s Law
!
!
 Fspring = !kx
This force is not constant – it changes with changing deformation (x) in a spring.
! ! x2
2
Fid
! x = ! ( "kx ) dx = " 12 kx
x2
Wspring =
x1
x1
x2
x1
(
)
= " 12 kx22 " " 12 kx12 = 12 kx12 " 12 kx22
Block on Incline Collides with Spring: A 12 kg block is released from rest on a 30°
frictionless incline. Below the block is a spring that can be compressed 2.0 cm by a
force of 270 N. The block momentarily stops when it compresses the spring by 5.7 cm.
From the given data, determine the spring constant: k =
Fspring
x
=
270N
= 13, 500 N m
0.02m
(a) How far does the block move down the incline from its rest position to this stopping
point?
W = !K = 12 m v 2f " 12 m v02 = 0
!"
!"
=0
=0
Wgravity = mg sin 30°!x
Wspring = 12 k x12 ! 12 kx22
!"
 W = Wgravity + Wspring = mg sin 30°!x " 12 kx22 = 0
=0
1
kx22
(13500 N m ) ( 0.057m ) = 0.373m = 37.3cm
!x =
= 2
mg sin 30° (12kg ) ( 9.8 m s2 ) sin 30°
1
2
2
(b) What is the speed of the block just as it touches the spring?
At the instant it just touches the spring, the block has moved (37.3 cm – 5.7 cm) = 31.6
cm along the ramp. Since the spring is not yet compressed, it doesn’t do any work, the
only work done is by gravity:
Wgravity = mg sin 30°!x = (12kg ) ( 9.8 m s2 ) ( sin 30° ) ( 0.316m ) = 18.6J
W = !K = 12 m v 2f " 12 m v02
!"
!"
=0
=?
18.6J =
1
2
(12kg ) v 2f
vf =

2 (18.6J )
= 1.76 m s
(12kg )
Let’s imagine a situation where a force is applied to an object and as a result, that
object moves.
!
Let F = ( Fx ) iˆ + Fy ĵ + ( Fz ) k̂
( )
!
And r = ( x ) iˆ + ( y ) ĵ + ( z ) k̂
!
!
When the object moves a very small distance ( dr ), we say: dr = ( dx ) iˆ + ( dy ) ĵ + ( dz ) k̂
!
!
The work done by this force is: W = ! F ( r )idr
r2
r1
How do we calculate this dot product?
!
!
F ( r )idr = Fx iˆ + Fy ĵ + Fz k̂ i dxiˆ + dyĵ + dzk̂ = Fx dx iˆiiˆ + Fy dy ĵi ĵ + Fz dz k̂i k̂
"$#
"$$#
"$$#
(
)(
)
r2
x
y2
z2
!
! 2
W = ! F ( r )idr = ! Fx dx + ! Fy dy + ! Fz dz
r1
x1
y1
z1
( )
( )
( )
=1
=1
=1