Download Chapter 6 – Work and Kinetic Energy

Chapter 6 – Work and Kinetic Energy
I.
Dot (or scalar) Product reminder


Given the vectors A and B :
 
 
Definition: A  B  A B cos 
Unit vectors:
iˆ  iˆ  1
ˆj  ˆj  1
kˆ  kˆ  1
iˆ  ˆj  0
ˆj  kˆ  0
kˆ  iˆ  0
ˆj  iˆ  0
kˆ  ˆj  0
iˆ  kˆ  0
 
A  B  Ax Bx  Ay By  Az Bz
II.
Work done by a Constant Force

Suppose a constant force F , acting at an angle , is exerted on an object that undergoes a

displacement s .

F


s
A.
Definition:
Work = Force in the direction of the displacement times the displacement;
W  F cos s  Fss  Fs cos  .
Or, Work = Force times the displacement in the direction of the force; or
Work = the magnitude of the Force times the magnitude of the Displacement times
the cosine of the angle between them.
 
W  Fs cos  Fss  F  s
6-1
B.
Units:
System
[Force]
[Displacement]
[Work]
SI
cgs
British
Engineering
C.
Examples:
1.
A 10 kg block is pulled 3 meters across a horizontal surface by a force of 40 N acting at an angle
of 37o. The coefficient of friction between the block and the surface is 0.25. Find the work done.
( WF , WN , Wmg , W f , and Wtotal or Wnet )
6-2
2.
A 10 kg block is pulled 3 m up a 37o incline by a force parallel to the incline at an acceleration of
2 m/s2. The coefficient of friction between the blcok and the incline is 0.25. Find
( WF , WN , Wmg , W f , and Wtotal or Wnet )
III. Work Done by a Variable Force
A.
If the force is not constant, then the definition of work done by a constant force can be used as an
approximation as long as the displacement is small. (Over a small displacement, the force does not
change very much and can be considered to be almost constant.)
 
W  F  s
This approximation becomes an equality in the limit as the displacement goes to zero, i.e.,
 
dW  F  ds  Fds cos  ,
or,
 
W  F  ds  Fds cos   Fs ds .



6-3
B.
Graphically, the work done by a force Fs moving an
object from s1 to s2 equals the area under the Fs curve
Fs
from s1 to s2 .
ds
s1
C.
s2
Work done by a Spring (Variable Force)
1.
Suppose we have a spring attached to a wall on the left and the spring is neither stretched nor
compressed. The position of the right end of the spring is called the equilibrium position (e.p.)
and labeled as x = 0. All displacements are measured from this position.
e.p.
k
 Fx
x =0
Fx
+x
When the spring is stretched to a displacement x by a force Fx to the right, the spring exerts a
force Fx to the left of the same magnitude. It is found that as long as the spring is not stretched
beyond its elastic limit, then the spring force is proportional to the displacement, i.e., Fx  x .
By introducing a proportionality constant an equation can be written as
Fs  Fx  kx , (Hooke’s Law),
where k is a constant characteristic of the spring and referred to as the spring or elastic constant.
The negative sign indicates that the force exerted by the spring and the displacement are in
opposite directions.
Units:
6-4
s
2.
Find the elastic (or spring) constant of the spring.
The value of the elastic constant k is found experimentally by hanging the spring vertically, then
adding weights and measuring the amount of stretch.
3.
D.
Find the work done by a spring (spring force) as the end of the spring is moved from position x1
(initial position) to x2 (final position).
If the force and displacement are expressed in terms of unit vectors, then the work done can be
written as
 
W  F  ds 

W
 F iˆ  F ˆj  F kˆ  dxiˆ  dyˆj  dzkˆ ,
x
y
z
 F dx  F dy  F dz   F dx   F dy   F dz .
x
y
z
x
y
z

Example: A force F  2 x 2 iˆ  3yˆj (in Newtons) acts on an object. Find the work done by this force
when the object moves from point (0, -1) to point (4, 5), where the distance is in meters.
6-5

Brain teaser: Suppose the force in the previous problem was F  2 y 2 iˆ  3xˆj . What other information
is needed to find the work? (Path?)
IV. Work Done by a Force Along a Curved Path

Suppose a force F is acting on an object and moving it along a curved path, as shown.

ds


F
A.

Method 1: The work done over the small length d s is
 
dW  F  ds  Fsds  F cosds  Fds cos .
To get the work done over the path, integrate


W  Fs ds  Fds cos  .
B.
Method 2: Set up x and y axes at the position of the object.
y

ds
x

F
6-6


Notice that both ds and F can be broken up into x and y components, and can be written as


ds  dxiˆ  dyˆj and F  Fxiˆ  Fy ˆj .
If this is done, then the work done is
 
W  F  ds  Fx dx  Fy dy .

C.


Example: A simple pendulum consists of a small ball of mass m suspended from a light string of
length L. The mass is pulled to the side until the string makes and angle of  with the vertical. The
ball is released from this position and swings downward along an arc due to the pull of gravity.
1.
Find the amount of work done by the gravitational force as the ball swings from its initial
position to the position it occupies when the string is vertical.
2.
Find the work done by the tension for this motion.
V.
Kinetic Energy and the Work-Energy Theorem
A.
A mass m is acted upon by external forces. Newton’s Second Law can then be written:

 F  ma .

When the total or net amount of work done by the forces is found, then the result is a useful
relationship call the “work-energy theorem.”
6-7
Derive the work-energy theorem starting with Newton’s Second Law:
B.
Kinetic Energy
Definition:
1
mv 2 = kinetic energy = K = “the ability or capacity of a body to do work because of its
2
motion.
Units:
C.
The work-energy theorem can be written as:
Wtotal  K2  K1  K
D.
Examples:
1.
A 10 kg block is pulled up a 37o inclined plane by a force F = 80 N parallel to the incline. The
coefficient of friction between the block and the incline is 0.20. The initial velocity of the block
is 2 m/s. How fast is the block traveling after the block moves 3 m up the incline?
6-8
2.
Find the coefficient of friction between the tires and the road for a car traveling at 30 m/s when
it comes to rest in 80 m as a result of the brakes being applied.
3.
A 10 kg block is attached to a horizontal spring with spring constant of 400 N/m. The block is
pulled 10 cm to the right of the equilibrium position and released from rest. The coefficient of
friction between the block and the surface is 0.20. Find the speed of the block as it passes
through the equilibrium position.
6-9
VI. Power = rate at which work is being done (also the rate at which energy is supplied, or the rate at
which energy is dissipated)
A.
Average Power:
Units:
Pav 
work done
W

time to do the work
t
joules
ft  lb
 Watt (W) , also 550
 1 horsepower (hp)  746 Watts (W)
sec
sec
 
W dW F  ds  
P  lim Pav  lim


 Fv
t 0
t 0 t
dt
dt
B.
Instantaneous Power:
C.
Example: A 2000 kg car accelerates uniformly from rest for 5 sec to a speed of 20 m/s.
1.
Find the total work done on the car in 5 sec.
2.
Find the average power needed to accelerated the car over the 5 sec.
3.
Find the instantaneous power needed by the car at time t = 5 sec.
6-10