Download Homework #4, Power

Homework #4, Power
1.
A 2.0-m-long pendulum is released from rest when the support string is at an
angle of 25° with the vertical. What is the speed of the bob at the bottom of the swing?
2.
An archer pulls her bowstring back 0.400 m by exerting a force that increases
uniformly from zero to 230 N. (a) What is the equivalent spring constant of the bow? (b)
How much work does the archer do in pulling the bow?
3.
(I) How long will it take a 1750-W motor to lift a 315-kg piano to a sixth-story
window 16.0 m above?
4.
(I) If a car generates 18 hp when traveling at a steady 88 km h , what must be the
average force exerted on the car due to friction and air resistance?
5.
(I) A 1400-kg sports car accelerates from rest to 95 km h in 7.4 s. What is the
average power delivered by the engine?
6.
(II) A driver notices that her 1150-kg car slows down from 85 km h to 65 km h in
about 6.0 s on the level when it is in neutral. Approximately what power (watts and
hp) is needed to keep the car traveling at a constant 75 km h ?
7.
8.
(II) How much work can a 3.0-hp motor do in 1.0 h?
A skier of mass 70 kg is pulled up a slope by a motor-driven cable. (a) How much
work is required to pull him 60 m up a 30° slope (assumed frictionless) at a constant
speed of 2.0 m/s? (b) What power must a motor have to perform this task?
9.
While running, a person dissipates about 0.60 J of mechanical energy per step per
kilogram of body mass. If a 60-kg person develops a power of 70 W during a race, how
fast is the person running? (Assume a running step is 1.5 m long.)
10.
The electric motor of a model train accelerates the train from rest to 0.620 m/s in
21.0 ms. The total mass of the train is 875 g. Find the average power delivered to the
train during its acceleration.
Solutions #4: Power
1.
The support string always lies along a radius line of the
circular path followed by the bob. This means that the
tension force in the string is always perpendicular to the
motion of the bob and does no work. Thus, mechanical
energy is conserved and (taking y  0 at the point of
support) this gives
1
2


m v2  12 m vi2  m g yi  yf  0  m g  L cosi  m g   L , or
v  2gL1 cosi  2 9.8 m s2   2.0 m
1 cos25
v  1.9 m s
2.
(a)
The equivalent spring constant of the bow is given by
k
(b)
Ff

xf
F  kx as
230 N
= 575 N m
0.400 m
From the work-energy theorem applied to this situation,

W nc  KE  PEg  PEs
   KE  PE
g
f

1


 PEs   0  0  kx2f   0  0  0
i


2
The work done pulling the bow is then
W nc 
3.
1 2 1
kxf   575 N m
2
2
 0.400 m 
2
 46.0 J
The work necessary to lift the piano is the work done by an upward force, equal in magnitude to the weight of the
piano. Thus W  Fd cos 0  mgh . The average power output required to lift the piano is the work done
divided by the time to lift the piano.
o
P
4.
W
t

mgh
t
 t
mgh
P

 315 kg   9.80 m

s 2 16.0 m 
1750 W
 28.2 s
The 18 hp is the power generated by the engine in creating a force on the ground to propel the car forward. The
relationship between the power and the force is given by
P
W
t

Fd
t
F
d
t
 Fv . Thus the force to
propel the car forward is found by F  P v . If the car has a constant velocity, then the total resistive force
must be of the same magnitude as the engine force, so that the net force is zero. Thus the total resistive force is
F  P v.
P 18 hp  746 W 1 hp 
F 
 5.5  102 N
v
 1m s 
 88 km h  

 3.6 km h 
also found by
5. The power is given by Eq. 6-16. The energy transformed is the change in kinetic energy of the car.

 1m s  
1400 kg   95 km h  

2
2

1
3.6 km h  
energy transformed KE 2 m  v2  v1 


P



time
t
t
2  7.4 s 
 6.6  104 W  88 hp
6. The energy transfer from the engine must replace the lost kinetic energy. From the two speeds,
calculate the average rate of loss in kinetic energy while in neutral.
2
 1m s 
  23.61m s
 3.6 km h 
v1  85 km h 
KE  12 mv22  12 mv12 
P
So
W

t
1.330  105 J
6.0 s
1
2
 1m s 
  18.06 m s
 3.6 km h 
v2  65 km h 
1150 kg  18.06 m s    23.61m s    1.330 10 5 J
2

2
 2.216  10 4 W , or 2.216  10 4 W
1 hp
 29.71 hp
 746
W
2.2  104 W or 3.0  101 hp is needed from the engine.
7.
Since
8.
(a)
P
W
, we have
3600 s 
 746 W 
6
1 hr  
  8.1 10 J

 1h 
 1 hp 
W  Pt  3.0 hp 
t
W nc  KE  PE , but KE  0 because the speed is constant. The skier rises a vertical distance
of y   60 m  sin30  30 m . Thus,


W nc   70 kg 9.80 m s2  30 m   2.06  104 J 21 kJ
(b)
The time to travel 60 m at a constant speed of
2.0 m s is 30 s. Thus, the required power input is
W nc 2.06 10 J
 1 hp 

  686 W  
  0.92 hp
t
30 s
 746 W 
4
Г
9.
N
t. We determine the number of steps per unit time by
 w ork perstep perunitm ass  m ass  # steps
w ork done

t

J step 
N
70 W   0.60
 60 kg  t  , giving

kg


Pow er=
or
t
N
= 1.9 steps s
t
,
The running speed is then
step  
x  N 
m 


distance traveled perstep   1.9
1.5
 2.9 m s





t  t 
s 
step 
Assuming a level track, PE f  PEi , and the work done on the train is
vav 
10.
W nc   KE  PE  f   KE  PE i

  0.875 kg  0.620 m s
 12 m v2f  vi2 
1
2
The power delivered by the motor is then
P
W nc
0.168 J

 8.01 W
t 21.0 10-3 s
2
 0  0.168 J
