Physics 203 – College Physics I
Department of Physics – The Citadel
Physics 203
College Physics I
Fall 2012
S. A. Yost
Chapter 6 Part 1
Work and Kinetic Energy
Physics 203 – College Physics I
Department of Physics – The Citadel
Announcements
Read Chapter 6 for next time. You can skip sec. 6-2.
Today we will discuss sec. 6.1, 6.3, and 6.10 on
Work, Kinetic Energy, and Power.
There is a problem set on these sections due
Thursday: HW3A
Next time, we will discuss sec. 6.4 – 6.9: Potential
Energy, Energy Conservation, Non-Conservative
Forces.
Problem set HW3B is due next Tuesday.
Physics 203 – College Physics I
Department of Physics – The Citadel
Newton’s Law and Orbits
First: an orbit problem…
Determine the mass of the sun using the properties
of Earth’s orbit. (You can treat it as circular.)
Newton’s Gravitational law:
Newton’s 2nd Law:
F = G Ms Me / R2.
F = Mea = Mev2/R
Uniform circular motion: v = 2pR / T
G Ms Me / R2 = Me (2p/T)2 R
Ms = (2p/T)2 R3 / G
Physics 203 – College Physics I
Department of Physics – The Citadel
Newton’s Law and Orbits
Ms = 4p2 R3/(GT2)
R = 1.50 × 1011 m
T = 1 year = 3.16 × 107 s
G = 6.67 × 10–11 Nm2/kg2
The numbers give Ms = 2.00 × 1030 kg
Physics 203 – College Physics I
Department of Physics – The Citadel
Effect of Force over Distance
Applying a force to a particle over distance
changes its speed in the direction of the
force:
vf2 – vi2 = 2ax
(1 dim)
Multiply by ½ m: ½ mvf2 – ½mvi2 = max
Use F = ma:
½ mvf2 – ½ mvi2 = Fx
Physics 203 – College Physics I
Department of Physics – The Citadel
The Work-Energy Theorem
Definitions: Kinetic Energy = K = ½ mv2.
Work = W = Fx.
Units: Work = N m = Joules (J).
Work-energy principle: The work done by the
net force on a mass causes a change in
kinetic energy:
DK = W
Physics 203 – College Physics I
Department of Physics – The Citadel
Work Requires Motion
Work is done only when there is
motion.
W = Fx requires both F and x to
be nonzero for W to be
nonzero.
You can push all day on a wall
and get very tired, but if it
doesn’t move, you did no work
on it.

F
Physics 203 – College Physics I
Department of Physics – The Citadel
Question
Suppose you apply a force Fp = 50 N to a box,
which causes it to move at a constant speed
through a distance of 10 m. What is the net
work done on the box?
A) 0.2 J
B) 500 J
C) 250 J
D) 0 J
Ff
Fp = 50 N E) 5 J
10 m
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
Suppose you apply a force Fp = 60 N to a box of
mass m = 15 kg initially at rest, with coefficient
of kinetic friction mk = 0.3.
Two forces act on the box:
Fp and Ff = mkmg.
Fp
Fp
x
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
1. How much work do you do on the box when it
moves a distance x = 12 m?
You do an amount of work
Wp = Fp x = (60 N)(12 m) = 720 Nm
= 720 J
Fp
The pushing force is in the
direction of motion, so
the work is positive.
x
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
2. How much work does friction do when you move
the box 12 m?
The force of friction is
Ff = – mk mg = – (0.3)(15 kg)(9.8 m/s2) = –44 N
The work done by friction is
Fp
Wf = Ff x = (–44 N)(12 m) = –528 J
x
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
3. What is the kinetic energy of the box after being
pushed 12 m?
The box was initially at rest, so the kinetic energy is the
net work,
K = Wp + Wf = 720 J – 528 J = 192 J
4. What is the speed of the box after being pushed 12 m?
The kinetic energy is K = ½ m v2 .
v = √ 2 K/m = √ 2(192 J) / 15 kg = 5.1 m/s
Physics 203 – College Physics I
Department of Physics – The Citadel
Orbit
A satellite of weight mg is in a near-Earth circular
orbit. How much work does gravity do on the
satellite during each orbit?
A) mgRe
B) 2pmgRe
Re
C) 0
Think of the work-energy principle: DK = W.
Physics 203 – College Physics I
Department of Physics – The Citadel
Work in More than 1 Dimension
Only the component of force in the direction
of motion does work.
y
W = (F cos q) x
→
F
θ
x
A force perpendicular to the motion does no
work!
It can’t change the speed, so it doesn’t affect
the kinetic energy. W = DK
Physics 203 – College Physics I
Department of Physics – The Citadel
High Dive via Work/Energy
v0
5.0 m
A diver jumps with initial speed
v0 = 1.4 m/s. At what speed does she
enter the water 5.0 m below?
Kf = K0 + W with K0 = ½ mv02,
Kf = ½ mvf
2.
Gravity is a constant force in the -y
direction, so W = - mg Dy = mgh.
The x motion does not affect the work.
vf
Physics 203 – College Physics I
Department of Physics – The Citadel
High Dive via Work/Energy
v0
Kf = K0 + W
5.0 m
½ mvf2 = ½ mv02 + mgh
vf = √v02 + 2gh
v0 = 1.4 m/s, g = 9.8 m/s2, h = 5.0 m.
vf = 10.0 m/s.
vf
Physics 203 – College Physics I
Department of Physics – The Citadel
Bowling Balls
A ball-feeder lifts balls up a 1 m long ramp to a
platform 0.5 m above the floor. How much
force must the feeder arm exert to lift a ball
weighing 50 N?
0.5 m
Enter the answer in Newtons.
Physics 203 – College Physics I
Department of Physics – The Citadel
Bowling Balls
The feeder does work Wp = Fp L = F × 1.0 m.
Gravity does work Wg = - mgh = -(50N)(0.5 m)
= - 25 N
Wp must be at least 25 J.
0.5 m
Fp must be at least 25 N.
Physics 203 – College Physics I
Department of Physics – The Citadel
Power
Power is the rate of doing work: P = W/t.
If the force F acts in the direction of motion, then
P = Fv
(instantaneous)
These are consistent because x = v t is the distance
traveled, so
P = F v = F x/t = W/t.
Physics 203 – College Physics I
Department of Physics – The Citadel
Bowling Balls
If the ball feeder is powered by a 5 W motor,
how many balls per minute can it lift, in
continuous operation?
0.5 m
Each ball required Wp = 25 J of work. The
motor supplies (5 W) (60 s) = 300 J in a
minute. That is enough energy to lift 12 balls.
Physics 203 – College Physics I
Department of Physics – The Citadel
Pulleys
How much force do I have to
pull with to lift a block of
mass M at a constant
speed?
Fp
Solve it using work and
energy.
Physics 203 – College Physics I
Department of Physics – The Citadel
Pulleys
Pulling a distance L with
force Fp does work
Wp = Fp L
The block moves up a
distance ½ L while lifted
with a force FLift = Mg.
Wp = FpL = WLift = ½ LMg
Fp = ½ Mg
Fp
L
½L
FLift
Physics 203 – College Physics I
Department of Physics – The Citadel
Work by a Spring
When a spring is
compressed or
stretched, there is a
restoring force given
by Hooke’s Law,
F = – k x.
k = spring constant.
F
L
x
x = -L
x=0
Physics 203 – College Physics I
Department of Physics – The Citadel
Work by a Spring
Suppose a ball is placed
in front of the spring.
If it is held at x = –L and
then released, how
much work does the
spring do on the ball
as it returns to its
equilibrium position,
launching the ball?
m
x = -L
x=0
x
Physics 203 – College Physics I
Department of Physics – The Citadel
Work by a Spring
The work done by a changing force is the average
force times the distance.
F
The force decreases from
kL down to 0 linearly, kL
so the average force
is F = ½ kL.
W=FL=½
W = ½ k L2
kL2.
0
-L
0
x
Physics 203 – College Physics I
Department of Physics – The Citadel
Spring Gun
Suppose a ball of mass
m = 250 g is pushed back
an distance L = 4 cm on a
spring of spring constant
k = 120 N/cm and released.
m
L
How much work is needed
to compress the spring?
L = 0.040 m
k = 12,000 N/m
W = ½ kL2 = 9.60 J.
Physics 203 – College Physics I
Department of Physics – The Citadel
Spring Gun
When the spring is released,
what is the launch speed
of the ball?
Work-Energy Theorem:
K = ½ mv 2 = W = 9.60 J.
m = 0.250 kg.
v = √2K/m = 8.76 m/s.
m
v