6. Work, Energy & Power
1. Work
2. Forces that Vary
3. Kinetic Energy
4. Power
What’s the speed of these skiers at the bottom of the slope?
Does the work of climbing a mountain depend on the route chosen?
No
Direct application of Newton’s law can be infeasible.
simple complicatedEnergy conservation to the rescue (Chap 7).
6.1. Work
Work W done on an object by a constant force F is
FW F r rF = displacement along direction of F.
Note: F need not be a net force.
1W Joule N m
cosW F x
FF x
Fr = force along direction of r .rF r
Example 6.1. Pushing a Car
The man pushes with a force of 650 N, moving the car 4.3 m.
How much work he does?
W F x 650 4.3N m 2.8 kJ
Example 6.2. Pulling a Suitcase
Woman exerts 60 N force on suitcase, pulling at 35 angle to the horizontal.
How much work is done if the suitcase is moved 45 m on a level floor?
xW F x
60 cos 35 45N m
2.2 kJ
x
xF = x cos
FF x
cosF x
cosF x
Work & the Scalar Product
Work is a scalar.
Scalar = quantity specified by a single number that is the same in every coordinate system.
Scalar has no direction.
Scalar (dot) product of vectors A & B :
cosA B A B is a scalar
cosW F r = angle between F & r
W F r
x x y yA B A B A
B
BA = B cos
cosA B A BAA B
BA B
x x y y z zA B A B A B 3-D
2-D
Example 6.3. Tugboat
Tug boat pushes a cruiser with force F = ( 1.2, 2.3 ) MN,
displacing the ship by r = ( 380, 460 ) m.
(a)Find the work done by the tugboat.
(b) Find the angle between F & r.
W F rx yF x F y
1.2 380 2.3 460MN m MN m
1510 MJ
cosF r
2 2x yF F F 2 2
1.2 2.3MN MN 2.59 MN
2 2r x y 2 2
380 460m m 597 m
1cosW
F r
1 1510
cos2.59 597
MJ
MN m 12
6.2. Forces that Vary
1
N
ii
W W
1
N
ii
F X x
1
limN
iN
i
W F X x
2
1
x
xW F x d x
2 1x xx
N
1
1
2iX x i x
1 1
1
2X x x
2 11
1
2N
x xX x N
N
2
1
2x x
Tactics 6.1. Integrating
inverseDerivative Indefinite Integral Antiderivative
1n
nd xn x
d x
ng x d x nd gx
d x
g f d x
d g
fd x
Example:
Since we have1
1
nxg
n
See Appendix A for integral table.
2 2
11
x xn
xxx d x g
1 12 1
1 1
n nx x
n n
2 1g x g x
1
1
nn x
x dxn
Stretching a Spring
0
xW F x d x 0
xk x d x
2
0
1
2
x
k x1
1
nn x
x dxn
21
2k x
Example 6.4. Bungee Jumping
Bungee cord is 20 m long with k = 11 N/m.
At lowest point, cord length is doubled.
(a) How much work is done on cord?
(b) How does work done in the last meter compare with that done in the 1st meter?
2 21 2 2 1
1
2W k x x
(a) 2 2111 / 20 0
2W N m m m 2.2 kJ
(b) 1st meter
2 2111 / 1 0
2W N m m m 5.5 J
Last meter
2 2111 / 20 19
2W N m m m 214 J
Example 6.5. Rough Sliding
Workers pushing a 180 kg trunk across a level floor encounter a 10 m region where floor
becomes increasingly rough.
There, k = 0 + a x2, with 0 = 0.17, a = 0.0062 m2 & x is the distance into the rough part.
How much work does it take to push the trunk across the region?
2
1
x
xW F x d x kF x m g
2
1
20
x
xW ax m g d x
2
1
30
1
3
x
x
m g x a x
3 30 2 2 0 1 1
1 1
3 3m g x a x x a x
32 21180 9.8 / 0.17 10 0.0062 10
3kg m s m m m
6.6 kJ
1
1
nn x
x dxn
Force & Work in 2- & 3- D
2
1
W dr
rF r r Line integral
2
1x yF d x F d y
r
r
2
1x y zF d x F d y F d z
r
r
2
1
x
xF d x 1-D
2-D
3-D
Work Done Against Gravity
Only vertical displacement requires work against gravity
W = m g h
W m g y
GOT IT? 6.2.
3 forces have magnitudes in N that are numerically equal to
(a) x, (b) x2, (c) x,
where x is the position in meters.
Each force moves an object from x = 0 to x = 1 m.
Note that each force has the same values at the end points, namely, 0 N & 1 N.
Which force does the most work?
Which does the least?
(c)
(b)
6.3. Kinetic Energy
net netW F d x dvm d x
dt
d xm dv
dt m v dv
2
1
v
net vW m v dv
2
1
21
2
v
v
m v 2 22 1
1 1
2 2m v m v
21
2m v
Kinetic energy: 21
2K m v
K is relative (depends on reference frame).
K is a scalar.
Work-energy theorem: netK W
Example 6.6. Passing Zone
A1400 kg car enters a passing zone & accelerates from 70 to 95 km/h.
(a) How much work is done on the car?
(b) If the car then brakes to stop, how much work is done on it?
2 22 1
1 1
2 2netW K m v m v
(a) 2 211400 95 / 70 /
2netW kg km h km h 22,887,500 /kg km h
21000
28875003600
mkg
s
222.8 kJ
b) 2 211400 0 95 /
2netW kg km h 26,317,500 /kg km h
21000
6,317,5003600
mkg
s
487.5 kJ
GOT IT? 6.3.
For each situation, tell whether the net work done on a soccer ball is
positive, negative, or zero.
Justify your answer using the work-energy theorem.
(a)You carry the ball to the field, walking at constant speed.
(b) You kick the stationary ball, starting it flying through the air.
(c) The ball rolls along the filed, gradually coming to a halt.
zero (K=0)
positive (K>0)
negative (K<0)
Energy Units
[energy] = [work] = J (SI)
CGS: 21 1 /erg g cm s 710 J
Other energy units:
eV (electron-volt): used in nuclear, atomic, molecular, solid state physics.
cal (calorie), BTU (British Thermal Unit): used in thermodynamics.
kW-h (kilowatt-hours): used in engineering.
See Appendix C
6.4. Power
Average power:W
Pt
(Instantaneous) power:0
limt
WP
t
dW
d t
power watt /W J s
Example 6.7. Climbing Mount Washington
A 55 kg hiker makes the vertical rise of 1300 m in 2 h.
A 1500 kg car takes ½ h to go there.
Neglecting loss to friction, what is the average power output for each.
Hiker:m g hP
t
255 9.8 / 1300
2 3600 /
kg m s mP
h s h 97W
21500 9.8 / 1300
0.5 3600 /
kg m s mP
h s hCar: 11 kW
W P t for constant power P
0limt
W P t
WP
t
2
1
t
tP dt general case
Example 6.8. Yankee Stadium
Each of the 500 floodlights at Yankee stadium uses 1.0 kW power.
How much do they cost for a 4 h night game, if electricity costs 9.5 ₵ / kW-h ?
W P t 500 1.0 4 $ 0.095 /kW h kW h $190
Energy and Society
2008: 4.71020 J or 1.51013 W
Power & Velocity
dW d F rd
dt F
rdWP
d t P F v
Example 6.9. Bicycling
Riding a 14 kg bicycle at a steady 18 km/h (5.0 m/s),
you experience a 30 N force from air resistance.
If you mass 68 kg, what power must you supply
(a)on level ground.
(b) up a 5 slope.
(a) airP F v 30 5.0 /N m s 150W
(b) air gP F F v sinairF m g v
230 14 68 9.8 / sin 5 50 /N kg kg m s m s 500W
Fair
Fg
vFair
v