Download Kinematics Multiples

Power
From Princeton Review Book
1. A force of 200 Newtons is required to keep an object sliding at a constant speed of 2
m/s across a rough floor. How much power is being expended to maintain this motion?
a. 50 Watts.
b. 100 Watts.
c. 200 Watts.
d. 400 Watts.
e. Cannot be determined from the information given.
* D. Using the formula which we don’t use often but which is really helpful:
P = F*v = (200N) (2 m/s) = 400 Watts.
2. A crane lifts a shipping crate that weighs 5,000 Newtons at a constant speed of 4 m/s.
At what rate is this crane doing work on the crate?
a. 1,250 Watts.
b. 2,000 Watts.
c. 4,000 Watts.
d. 10,000 Watts.
e. 20,000 Watts.
*E. Power = F*V = (5,000 N) (4 m/s) = 20,000 Watts.
3. If L, M and T denote the dimensions of length, mass, and time, respectively, what are
the dimensions of power?
ML
a. 2
T
*D.
c.
M 2L
T2
e.
M 2L
T3
L2 M
b.
T2
d.
L2 M
T3
Power 

Work ( force)( dis tan ce) (mass * acceleration)( dis tan ce)


Time
Time
time
(mass)( dis tan ce /(time) 2 )( dis tan ce) (mass)( dis tan ce) 2 ML2

 3
time
(time) 3
T
4. An engine provides 10 kW of power to lift a heavy load at constant velocity a distance
of 20 meters in 5 seconds. What is the mass of the object being lifted?
a. 100 kg.
b. 150 kg.
c. 200 kg.
d. 250 kg.
e. 500 kg.
*D.
P  Fv
(10,000Watts)  (mg )( d / t )
10,000Watts  m(10m / s 2 )( 20m / 5 sec)
m  250kg
From Old AP’s
5. (1974) Suppose the force required to tow a canal barge is directly proportional to
the speed. If it takes 4.0 horsepower to tow the barge at a speed of 2.0 mph, what
horsepower is required to move the barge at a speed of 6.0 mph?
a. 8.0 horsepower.
b. 12 horsepower.
c. 24 horsepower.
d. 32 horsepower.
e. 36 horsepower.
*E. This one is really tricky. They give you information about force and speed, but ask
about power, so you need to combine a few formula:
Say F  bV since you are given that F is proportion al to V.
Then, Power : P  Fv  (bv)v  bv 2
Solve for b :
4hp  b(2mph) 2
so b  1.
Now recalculat e with second speed :
P  (1 hp/mph 2 )(6mph) 2  36hp
You could also reason tha t since power vari es as velocity squared, if you triple the velocity
the power will increase by a factor of 9.
6. A weight lifter lifts a mass m at constant speed to a height h in time t. What is the
average power output of the weightlifter?
a. mg
b. mh
c. mgh
d. mght
e. mgh/t
* E. Power 
Work Fd cos  (mg )h cos(0)


time
t
t
7. (1984) A person pushes a box across a horizontal surface at a constant speed of 0.5
m/s. The box has a mass of 40 kg, and the coefficient of sliding friction is 0.25. The
power supplied to the box by the person is:
a. 0.2 Watts.
b. 5 Watts.
c. 50 Watts.
d. 100 Watts.
e. 200 Watts.
* C. The fact that the box is traveling at a constant speed is a clue that the box is in
equilibrium and thus the force exerted by the person is the same as the force of friction.
P  Fv  ( Friction )V  ( mg )(V )  (.25)( 40kg)(10m / s 2 )(. 5m / s )
 50Watts
8. (1993) During a certain time interval, a constant force delivers an average power of 4
Watts to an object. If the object has an average speed of 2 m/s and the force acts in the
direction of motion of the object, the magnitude of the force is:
a. 16 N.
b. 8 N.
c. 6 N.
d. 4 N.
e. 2 N.
* E. This is pretty much plug and chug into the power formula.
P  Fv
4Watts  F (2m / s)
F  2N
9. A child pushes horizontally on a box of mass m which moves with constant speed V
across a horizontal floor. The coefficient of friction between the box and the floor is .
At what rate does the child do work on the box?
a. mgV
b. mgV
c. V/ mg
d. mg/V
e. mV2
* A. This is a classic problem. They are testing two things:
that if the box is traveling at a constant speed, the force of the child equals friction.
The plug-and-chug formula for power.
P  Fv  ( friction)V  mgV
10. A student weighing 700 Newtons climbs at constant speed to the top of an 8 m
vertical rope in 10 seconds. The average power expended by the student to overcome
gravity is most nearly:
a. 1.1 Watts.
b. 87.5 Watts.
c. 560 Watts.
d. 875 Watts.
e. 5600 Watts.
* C. If the student climbs at a constant speed, the force she exerts is equal to her weight.
Her speed is 8m/10s = .8 m/s.
P  FV  (mg )(V )  (700 N )(.8m / s)  560Watts
11.
(2004, 54%)
A 1000 Watt electric motor lifts a 100 kg safe at constant velocity. The vertical distance
through which the motor can raise the safe in 10 seconds is most nearly:
a. 1 meter.
b. 3 meters.
c. 10 meters.
d. 32 meters.
e. 100 meters.
*C.
P
Work
time
1000Watts 
h  10meters
mgh
(100 kg) (10 m/s 2 )( h)

t
10 sec
12.
(2004B, 40%)
A constant force of 900 N pushes a 100 kg mass up the inclined plane shown above at a
uniform speed of 4 m/s. The power delivered by the 900 N force is most nearly:
a. 2160 Watts.
b. 2700 Watts.
c. 2880 Watts.
d. 3600 Watts.
e. 4000 Watts.
*D
Since the force is pushing in the same direction as the motion, we don't have to break the
force vector down.
P  Fv  (900N) (4 m/s)  3600 Watts