Work don by a varying force
F
ΔW=FΔx
W   W   F ( x)x
a
b
Work on curved trajectories:
bx
xa
b
a
a
W   dW   F ( x)dx


W   F ( x, y, z )dl
b
a
dl
F
x
b
Example: Find the work done by gravity on a pebble of mass m as
it rolls from the rim to the bottom of a spherical bowl of radius R.
rim
W 
top
θ
dl = R dθ
mg
bottom

R
 = /2
 =0
 mgR
θ
bottom
mg  dl
mg Rd  cos

 /2
0
cos d 
 mgR sin 
 /2
0
 mgR
bottom
R


W   mgdl   m 0iˆ  gˆj dl x iˆ  dl y ˆj   mgdl y  mgR

bottom
top
top


0
 
WN   Ndl  0
W = ΔKE
mgR  12 mv 2
v  2 gR
Example: Car going down a hill. How much work is done by gravity
on a car which coasts down a hill of height h?
P2
 
W   F  dl
y
P1

F  mgjˆ
x
But what is the shape of the hill??
P2
Δr
P1
d
Ha! We don’t need it. Because the force is constant we can take it outside
of the integral:
P2
 
 P2 


W   F  dl  F   dl  F  r  ( mgjˆ)  ( diˆ  hjˆ)  mgh
P1
P1
The work by gravity does not depend on the shape of the hill
(i.e., on the path), only the drop h.
h
Example: You are speeding down the road without a care in the world and
just as you are approaching a sharp curve in the road above a scenic
overlook you notice the your brakes are soft due to leaking brake fluid.
From past experience you estimate that you have about 2 s of braking left.
Before the dangerous curve there is a dip in the road. Which is the best
place to use your limited brake? You must slow down before the curve or
you will become part of the scenery!
Not D!!!!
A
Decrease K
as much as
possible
B
Need
large|W|
Dip
Apply force
during long
distance
C
D
For fixed
time (2 s),
longest
distance
when large
speed
Largest speed is
at the dip
(work by gravity
increases K)
Power
Units:
dW
P
dt
 
P  F|| v  F  v
dW F|| dl
P

dt
dt
If
1W=1J/s
Horsepower:
1 hp = 746 W
 
P  F|| v  F  v
F||  const
Example: A force of 4 N acts on a 12 kg body initially at rest. Compute
a) the instantaneous power due to the force at the end of the third second;
b) the average power due to the force at the end of the third second
F  4N
m  12kg
v0  0
t  3s
P?
P ?
a)
a  F /m
v  at  Ft / m
P  Fv  F 2t / m
b)
v  12 v
P  Fv  12 Fv  12 P  2W
P  4 N  3s  / 12kf   4W
2
Example: Two elevators A and B carry each a load of mass m from the
first floor to the third floor of a building at constant speeds, but A is twice
as fast as B. 1) Compare work done by the cable tension (ie, the energy
produced by the engine). 1) Compare power of two elevators.
T
Δx
WTA  WTB
1) W=TΔX=mg ΔX
WT
 Tv
2) PT 
t
mg
t B  2t A
v A  2vB
PTA  2PTB
Example: Let’s say you load 500 kg of bricks by hand in 30 minutes, and
the same amount by forklift in 10 minutes. The pickup bed is 1.0 m high.
What is the power in each case?
Work is the same:
Phand
m
W  Fd  ( 500kg)( 9.8 2 )( 1.0m)  4900 J
s
W
4900 J


 2.7 W
t hand 30 min ( 60 s / min )
Plift 
W
 3Phand  8.2 W
t lift
Hook’s Law (elastic forces)
F=-kx
Fext
F
Fext
Fext
Fext
Fext=-F=kx
PE  W  Wext  12 Fext x  12 (kx) x  12 kx2
Wext
xi=0
xf
x
PE  12 kx2
PE  W  Wext  12 kx2f  12 kxi2
Fext
Wext
xi
xf
x
xf
xf
xi
xi
Wext   Fext ( x)dx   kxdx  12 kx 2f  12 kxi2
Example: A box of mass m = 25 kg slides on a horizontal frictionless
surface with an initial speed v0 = 10 m/s. How far will it compress the
spring before coming to rest if k = 3000 N/m?
m = 25 kg
v = 10 m/s
k = 3000 N/m
x
work-kinetic energy theorem: W = ΔKE
W   12 kx2
1
KE  0  mv 2
2
1 2 1
kx  mv 2
2
2
mv 2
(25 kg)(10 m/s)2
x 

 0.91 m
3000 N/m
k
A.
0.50 m
B.
0.63 m
C.
0.75 m
D.
0.82 m
E.
0.91 m