How can I calculate the work done by weight on a crate sliding down an incline?

In summary, the question is asking for the work done on a crate by different forces as it is pushed up a frictionless incline. The worker exerts a force of 209N parallel to the incline as the crate slides 1.5m. The book lists the answer for the work done by the worker as 314J, calculated by multiplying the force and the vertical displacement. This excess work goes into the kinetic energy of the crate.
  • #1
kdinser
337
2
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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  • #2
Once set in motion, is the crate moving with constant velocity (i.e., zero acceleration)?
 
  • #3
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
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
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
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: )
 

1. What is the definition of "Work Done by Weight"?

The work done by weight is a measure of the amount of energy expended when an object is moved against the force of gravity. It is calculated by multiplying the weight of an object by the distance it is moved vertically.

2. How is work done by weight different from work done by force?

Work done by weight only takes into account the vertical movement of an object against the force of gravity, while work done by force can involve movement in any direction against any type of force.

3. What is the formula for calculating work done by weight?

The formula for work done by weight is: W = mgd, where W is work, m is the mass of the object, g is the acceleration due to gravity, and d is the vertical distance the object is moved.

4. Can the work done by weight ever be negative?

Yes, the work done by weight can be negative if the object is moved in the opposite direction of the force of gravity. This means that the object is losing potential energy and the work done is considered negative.

5. How is work done by weight related to potential energy?

Work done by weight is directly related to potential energy. As an object is moved vertically against the force of gravity, it gains or loses potential energy depending on the direction of the movement. The work done by weight is equal to the change in potential energy of the object.

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