Work Done by Weight problem

  • Thread starter kdinser
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I must be missing something pretty basic here since I solved a similar problem earlier in the chapter with no problems.

The Question:
to push a 25.0kg crate up a frictionless incline, angled at 25.0 degrees to the horizontal, a worker exerts a force of 209N, parallel to the incline. As the crate slides 1.5m, how much work is done on the crate by (a) the workers applied force (b) the weight of the crate (c) the normal force (d) what is the total work done on the crate?

(a) is the one that is giving me problems. The book lists the answer as 314J

The change in d is .63 m given by 1.5 sin 25
The work done by gravity is m*g*d=-154 J

This is where I'm running into problems, shouldn't the negative work done by gravity=the work done by the worker?

A similar problem earlier in the chapter had a crate being pulled up the ramp by a rope, I don't see how that problem is any different then this one other then the initial info given. I applied the same technique to solving both problems and came up with the right answer for one and the wrong answer for the other. Thanks for any help.
 
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Answers and Replies

  • #2
robphy
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Once set in motion, is the crate moving with constant velocity (i.e., zero acceleration)?
 
  • #3
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The problem doesn't say, but it also doesn't say that it stopped. I just looked at the problem that is similar to this one and it does expressly say that it moves up to a certain height and stops.

Thanks robphy, that might be piece of info that I've been overlooking, I'll try again from that angle.
 
  • #4
Doc Al
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kdinser said:
... a worker exerts a force of 209N, parallel to the incline. As the crate slides 1.5m, how much work is done on the crate by (a) the workers applied force
....
(a) is the one that is giving me problems. The book lists the answer as 314J

The change in d is .63 m given by 1.5 sin 25
The work done by gravity is m*g*d=-154 J
The problem is straightforward. You are given the force and the displacement and asked to find the work. Well... work = Fd, right?

For some reason you are calculating the vertical displacement. That's not needed for part a. (Neither is the work done by gravity.)
 
  • #5
Gokul43201
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I guess you are being asked for the work done by the worker against gravity, so that's the product of the applied force and the vertical displacement. So, W = 209N * 1.5 sin(25)m = 314 J

Since the worker does more work than gravity, the excess goes into the KE of the crate, ie. it speeds up (unless there's friction, which i think there isn't).
 
  • #6
Doc Al
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nope

Gokul43201 said:
I guess you are being asked for the work done by the worker against gravity, so that's the product of the applied force and the vertical displacement. So, W = 209N * 1.5 sin(25)m = 314 J
That's not what's being asked. See my previous post. (And check those numbers! :uhh: )
 

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