Physics I
Class 10
Work and Kinetic Energy
Rev. 14-Feb-03 GB
10-1
Work
 Work is a measure of the energy that a force puts into (+) or
takes away from (–) an object as it moves.
 We will see that work is a useful way to solve problems where
the force on an object is a known function of position.
 Example: the force of an object connected to an ideal spring:


F  k x

x
where is the displacement from equilibrium. (Hooke’s Law)
10-2
Work for Constant Force
 
W  Fd
 
W  F d cos( )
10-3
Vector Dot Product
F

d
If you know lengths and angle:
 
W  F d cos( )
If in opposite directions:
W  Fx d x  Fy d y  Fz d z 
 
W Fd
 
W  F d
If at right angles:
W0
If you know components:
If in the same direction:
10-4
Work for Variable Force
W   F ( x )dx
xf
xi
(This is the version for one dimension.)
10-5
Work-Kinetic Energy Theorem
 Net work is done on an object by the net force:
xf
Wnet   Fnet dx
xi
 Kinetic energy defined for an object:

2
2
2
2
K  12 m v  12 m v x  v y  v z

 Work-Kinetic Energy Theorem: (without proof)
K f  K i  Wnet
10-6
Class #10
Take-Away Concepts
1.
Work is a measure
 ofenergy added to (+) or taken away (–).


W  F  d  F d cos( ) (constant force)
xf
W   F dx
(variable force, 1D)
xi
2.
Vector dot product defined.
3.
Kinetic Energy:
5.
6.
7.
Positive net work means an object’s K.E. increases (speeds up).
Negative net work means an object’s K.E. decreases (slows down).
Zero work means an object’s K.E. stays constant (constant speed).

2
2
2
2
K  12 m v  12 m v x  v y  v z
K f  K i  Wnet
4. Work-Kinetic Energy Theorem:

10-7
Class #10
Problems of the Day
1.__ Two bodies of unequal mass, placed at rest on a frictionless surface,
are acted on by equal horizontal forces over equal distances. As each
mass crosses the “finish line” (not necessarily at the same time), the
body of greater mass will have:
F
F
A)
B)
C)
D)
E)
The greater speed.
The same kinetic energy as the other body.
The greater kinetic energy.
The smaller momentum.
The same momentum as the other body.
10-8
Answer to Problem 1 for Class #10
The answer is B.
The work-kinetic energy theorem says that the change in kinetic
energy is equal to the net work done. The net work done is the same
for both objects (force dot displacement), therefore both objects have
the same kinetic energy when they cross the finish line.
The larger object has greater momentum (magnitude) than the
smaller object. Why? Hint: Will the same force act for a longer time
or a shorter time on the larger object? What does that say about the
impulse on the larger object?
10-9
Class #10
Problems of the Day
2. As an object with a mass of 3.2 kg moves from X = +3.0 m,
Y = +2.0 m to X = +6.0 m, Y = –2.0 m, it experiences a constant
force with X component of +4.0 N and Y component of +3.0 N.
How much work does the force do on the object?
10-10
Answer to Problem 2 for Class #10
The answer is zero.
It is important to remember that work is a scalar, not a vector.
The mass of the object is irrelevant.
For this problem, it is easier to use the component form of the dot
product, not the magnitude-angle form. (You will get the same
answer if you do it correctly, but it is more difficult.)
 
W  F  d  Fx d x  Fy d y  4  3  3  (4)  12  12  0
10-11
Activity #10
Work-Kinetic Energy Theorem
Objective of the Activity:
1.
Use LoggerPro to study the Work-Kinetic Energy
Theorem for a one-dimensional case.
10-12
Class #10 Optional Material
Work-K.E. Theorem Proof
xf
Wnet
xf
xf
xf
dv
  Fnet dx   m a dx  m  a dx  m 
dx
dt
xi
xi
xi
xi
Now we work on the integrand using calculus rules:
dv dv dx
dv 1 d 2
v 

v

dt dx dt
dx 2dx
Putting that back into the integral:
xf


1
d 2
v dx  1 m v f 2  vi 2  K f  K i
Wnet  m 
2 xi d x
2
10-13