Chapters
1–6
43
Power
Power is the rate of doing work, or the rate of change of
mechanical energy due to an applied force.
Unit: 1 watt (W) = 1 J/s
W Fs
s
Power, P = =
= F × = Fv
t
t
t
P = Fv
44
6.58
A 300 kg piano is being lifted by a crane at a steady speed to a
height of 10 m. The crane produces a steady power of 400 W.
How much time does it take to lift the piano?
Power P = Fv
(F = force to lift piano)
F
a=0
and F = mg = 300g (lifted at constant speed)
So P = mgv and v = P/(mg)
(400 W)
v=
= 0.136 m/s
(300 kg) × g
mg
h
10 m
Takes time =
= 73.5 s
v 0.136 m/s
45
6.62
A 1900 kg car travels up and down a hill at a constant
speed of 27 m/s. The force of air resistance and friction is
of the same magnitude in both directions.
Going up the hill, the car’s engine needs to produce 47 hp (35 kW)
greater power to maintain speed than it does going down the same
hill.
What is the angle of inclination of the hill?
v = 27 m/s
Fr
q
v = 27 m/s
Going up
Fr
q
Going down
46
v = 27 m/s
Power developed
by engine = P
l
Fr
l sin q
q
Going down
Work-energy theorem:
Work done by non-conservative forces (all but gravity)
= change in mechanical energy
That is, Wnc = DKE + DPE
and DKE = 0
In a time t the engine does work Pt and the car travels a distance
l = vt along the road
So, Wnc = (Pt − Fr l) = −mgl sin q
Or, (Pl/v − Fr l) = −mgl sin q
(DPE = −mgl sin q)
→ P/v = Fr − mg sin q
47
v = 27 m/s
Fr
Power developed
by engine = P!
q
Going up
When the car travels a distance l up the slope:
DPE = +mgl sin q and DKE = 0
So Wnc = P!l/v − Fr l = mgl sin q
and P!/v = Fr + mg sin q
P/v = Fr − mg sin q
→ P" − P = v(2mg sin q)
P! − P
So sin q =
2mgv
48
P! − P
sin q =
2mgv
P! − P = 35, 000 W
m = 1900 kg
v = 27 m/s
So q = 2◦
49
Other Forms of Energy
There are many forms of energy:
• Electrical
• Elastic (eg energy stored in a spring)
• Chemical
• Thermal
• Nuclear
Energy is conserved overall:
– energy may be converted from one form to another, but the
total amount of energy is conserved.
50
Work done by a variable force
Example: compound bow
– a number of pulleys and
strings
• maximize the energy
stored in the bow for
finite effort
• reduced force with
bow fully drawn.
51
Force required to pull
back the bowstring
Work done in drawing the
bow:
• split up the work into
segments
W ! (F cos q)1Ds1 + (F cos q)2Ds2 + . . .
Becomes exact as the segments Ds
become vanishingly narrow.
The work is then the area under
the curve – count the squares,
multiply by area of each.
52
Work done in drawing the bow
Number of squares under
the curve 242.
Area of each square is
(9 N)×(2.78×10-2 m) = 0.25
N.m = 0.25 J.
So, work done is
W = 242 × 0.25 = 60.5 J
53
6.64
Work done = area under triangular curve
1
= × (base) × (height)
2
W = 0.5 × (1.6 m) × (62 N) = 49.6 J
54
6.67
A force is applied
to a 6 kg mass
initially at rest.
a) How much
work is done by
the force?
b) What is the
speed of the mass
at s = 20 m?
a) Work done = area under the force-displacement curve
1
W = × (10 m) × (10 N) + (20 − 10 m)(10 N) = 150 J
2
55
b) What is the speed of the mass at s = 20 m?
Wnc = DKE + DPE = mv2/2 + 0 = 150 J
v=
!
2Wnc/m =
!
2 × 150/6 = 7.07 m/s
56
Summary
In absence of non-conservative forces:
Conservation of energy: E = KE + PE = constant
When non-conservative forces are present:
Work-energy theorem: Wnc = DKE + DPE
Power = rate of doing work (J/s = W)
P = Fv
Work done by a variable force = area under the force vs
displacement curve
57