work and power


by jensson
Tags: power, work
jensson
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#1
Feb25-11, 05:22 PM
P: 25
1. The problem statement, all variables and given/known data

A 7.5 kg box is being lifted by means of a light rope that is threaded through a single, light, frictionless pulley that is attached to the ceiling.

a) If the box is being lifted at a constant speed of 2.0 m/s, what is the power delivered by the person pulling on the rope?

b) If the box is lifted, at constant acceleration, from rest on the floor to a height of 1.5m above the floor in 0.42 s, what average power is delivered by the person pulling on the rope?


2. Relevant equations

k = 1/2 m v^2 (velocity vector)

d = initial velocity(t) + 1/2at^2

v = initial velocity + at


3. The attempt at a solution

My answer for a is 15 Joules. I basically just plugged in the numbers for this one, since we are only going in the y direction I didn't bother writing ( 0i + 2.0j)

k = (1/2)(7.5)(2.0)^2 = 15 Joules



My answer for b is 190 Joules. For this one I found the acceleration and then the velocity and plugged in those numbers.

1.5 = 1/2a(0.42)^2
a = 17 m/s^2

v = 17(0.42) = 7.1 m/s

k = 1/2 (17) (7.1)^2 = 190 Joules



Can anyone tell me if these answers are correct?
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collinsmark
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#2
Feb25-11, 09:33 PM
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P: 1,842
Hello jensson,
Quote Quote by jensson View Post
My answer for a is 15 Joules. I basically just plugged in the numbers for this one, since we are only going in the y direction I didn't bother writing ( 0i + 2.0j)

k = (1/2)(7.5)(2.0)^2 = 15 Joules
Sorry, but no.

You have found the box's kinetic energy, but that's not what the question is asking for. The question is essentially asking for the box's change in energy (both potential and kinetic), per unit time. That's a measure of power not energy. And power has units of Watts in the SI system of units, not Joules.

Let me give you an example (which isn't quite related to this problem). Suppose that the kinetic energy of a ball is 100 J. Ten seconds later it has a kinetic energy of 300 J, and suppose its potential energy is unchanged. The change in kinetic energy is 300 - 100 = 200 J over those 10 seconds. So the average power applied to the ball over those 10 seconds is (200 J)/(10 s) = 20 W.

Now back to this problem (part a). The block starts out at 2.0 m/s and keeps going indefinitely at 2.0 m/s. So, over a period of time, say 1 second (the block is still moving at 2.0 m/s), what is its change in kinetic energy?

But that's not the end of the problem. The potential energy of the block is also changing. This is because the block's height is changing. So, over some period of time, say 1 second, what is the change of the block's potential energy?
My answer for b is 190 Joules. For this one I found the acceleration and then the velocity and plugged in those numbers.

1.5 = 1/2a(0.42)^2
a = 17 m/s^2

v = 17(0.42) = 7.1 m/s

k = 1/2 (17) (7.1)^2 = 190 Joules
Ok, you found the block's kinetic energy at the end of the 0.42 s period. What is the block's change in kinetic energy over this period?

But you're not done yet. What is the block's potential energy at the start? What is the block's potential energy at the end of the 0.42 second period? So what is the change in potential energy?

What is the block's change in total energy (potential + kinetic) over the 0.42 second period? What's the average power? That's the average power exerted by the person pulling on the rope.

[Edit: By the way, there is another, equivalent method that you can use for both parts of this problem. Look in your textbook/coursework to find a formula that relates power to force and velocity. For part b, you can find the average power by using the average velocity.]


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