Problem #1
What power is required to pull a 5.0 kg block at a steady speed of 1.25 m/s? The coefficient of friction is 0.30.Answer:
The power required to move the block at constant speed is P = Fv. We are given v, the speed of the block. To get F, a force, we draw a FBD and apply Newton's Second Law,
The second equation gives N = mg and we know fk = μkN, so fk = μkmg.
Therefore, the applied force is F = μkmg.
Thus the power is
P = μkmgv
P = (0.3)(5 kg)(9.81 m/s2)(1.25 m/s) = 18.4 Watts
Problem #2
A 7500 W engine is propelling a boat at 12 km/h. What force is the engine exerting on the boat? What force and how much power is water resistance exerting on the speedboat?
Answer:
First we convert the velocity to SI units,
12 km/h × (1000 m)/km × (1 h)/(3600 s) = 3.333 m/s .
We know P = Fv, so
F = P/v = 7500 W / 3.333 m/s = 2250 N
By Newton's Third Law, the water is exerting 2250 N in the reverse direction. It is also removing 7500 W of power which is going into increasing the kinetic energy of the water.
Problems #3
A 3.0 hp engine pulls a 245 kg block at constant speed up a 12.0 m 30.0° incline. How long does this take? Ignore friction.
Answer:
First, 3.0 hp × 746 W/hp = 2238 W.
Next, the work done by the engine must equal the work done by the weight, from Newton's Third Law. The work done by gravity is
Wgrav = mgh = mgLsinθ = 1.442 × 104 J
Since power is defined as work over time, P = W/t, the time it takes is
t = W/P = (1.442 × 104 J)/(2238 W) = 6.44 s
Problem #4
A 700N marine in basic training climbs a 10.0 m vertical rope at a constant speed of 8.00 s. What is his power output?
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| Fig.1 |
Marine is shown in Fig. 1. The speed of the marine up the rope is
v = d/t = 10.0 m/8.00 s = 1.25 m/s
The forces acting on the marine are gravity (700 N, downward) and the force of the rope which must be 700N upward since he moves at constant velocity. Since he moves in the same direction as the rope’s force, the rope does work on the marine at a rate equal to
P = dW/dt = F · v = Fv = (700 N)(1.25 m/s) = 875W
(It may be hard to think of a stationary rope doing work on anybody, but that is what is happening here.)
This number represents a rate of change in the potential energy of the marine; the energy comes from someplace. He is losing (chemical) energy at a rate of 875W.
Problem #5
A 10.0 kg mass is dropped from a tall building. During the first second of the fall, what was the average power exerted by gravity? What was the average power exerted by gravity during the first 5.00 seconds of the fall?
Answer:
You need to first figure out how far the ball falls in 1.0 second and in 5.0 seconds…
y = y0 + v0yt + ½ gt2
but, v0y = 0.
Δy = ½ (9.8 m/s2)(1 s)2 = 4.9 m (during first 1.0 second)
Δy = ½ (9.8 m/s2)(5 s)2 = 123.5 m (during first 5.0 second)
Then, average power exerted by gravity during the first 5.00 seconds of the fall is
Pave = ΔE/t = mgy/t = (10 kg)(9.8 m/s2)(4.9 m) = 480 W (during first 1.0 s)
Pave = ΔE/t = mgy/t = (10 kg)(9.8 m/s2)(123.5 m) = 2400 W (during first 5.0 s)
Problem #6
Water flows over a section of Niagara Falls at a rate of 1.2 × 106 kg/s and falls 50 m. How many 60 W bulbs can be lit with this power?
Answer:
Whoa! Waterfalls? Bulbs? What’s going on here?? If a certain mass m of water drops by a height h (that is, Δy = −h), then gravity does an amount of work equal to mgh. If this change in height occurs over a time interval Δt then the rate at which gravity does work is mgh/Δt.
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| Fig.2 |
Wgrav = mgh = (1.2 × 106 kg)(9.80 m/s2)(50 m) = 5.88 × 108 J .
This occurs every second, so gravity does work at a rate of
Pgrav = mgh/Δt
= 5.88 × 108 J/1 s
Pgrav = 5.88 × 108 W
As we see later, this is also the rate at which the water loses potential energy. This energy can be converted to other forms, such as the electrical energy to make a light bulb function. In this highly idealistic example, all of the energy is converted to electrical energy. A 60 W light bulb uses energy at a rate of 60 J/s = 60 W. We see that Niagara Falls puts out energy at a rate much bigger than this! Assuming all of it goes to the bulbs, then dividing the total energy consumption rate by the rate for one bulb tells us that
N = 5.88 × 108 W/60 W
N = 9.8 × 106 bulbs can be lit.
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