# Concept of 'work done' on a smooth slope (high school-level physics)

1. ### question dude

80
Say you have an object on a smooth friction-less slope, and a force 10N (acting parallel to the slope) is applied to it to move it at a distance 5m up the slope.

Work done by the force is: force*distance. In this case, 10*5 = 50 joules of energy

Does all that energy get transferred to gravitational potential energy?

2. ### Simon Bridge

15,294
Re: Concept of 'work done' on a smooth slope (high school-level physic

Yes - provided the motion has constant speed.

3. ### question dude

80
Re: Concept of 'work done' on a smooth slope (high school-level physic

but this will only happen if the object's weight component (acting down the slope) is equal to the 10N applied force, right?

4. ### Simon Bridge

15,294
Re: Concept of 'work done' on a smooth slope (high school-level physic

That's right - for a constant velocity, the forces have to cancel out.
If the object accelerates then some of the work goes into extra kinetic energy.

5. ### question dude

80
Re: Concept of 'work done' on a smooth slope (high school-level physic

Heres what I'm a bit confused, anything that moves would have kinetic energy. So even if the object is moved at constant velocity, it would still have kinetic energy

and if it has kinetic energy, then how can it be true that all of the 'work done' by the 10N force gets transferred to gravitational potential energy?

also, what do you mean by 'extra kinetic energy' for when an object accelerates?

6. ### Philip Wood

1,176
Re: Concept of 'work done' on a smooth slope (high school-level physic

The thing about the object's KE is that you can make it as small as you like.

As usual, simple calculations aid understanding... Suppose the object had a mass of 2 kg, and that it was acceptable to move it 5 m up the slope in a time of 1000 s (rather long!) at a steadily increasing speed. The body's mean speed will then be $0.005 m s^{-1}$, so its final speed will be $0.01 m s^{-1}$, and its final KE will be $1\times 10^{-4} J$. Compare with the gain in grav PE, which is roughly 50 J. [A slope of 30° and a vertical height of 2.5 m is consistent with your figures and mine, taking g as $10 m s^{-2}$.]

Another thing to be clear about is that once an object is moving at a steady speed, it no longer needs work doing on it to keep it moving at that speed. For example, in the previous case we could choose to apply a larger excess (accelerating) force in the first few seconds, to get the object up to speed ($0.01 m s^{-1}$), and then we'd not need any excess force for the rest of the time. In this way we'd get the body 5 m up the slope in just over half the previous time, without using any more energy, as we'd have given the body no more KE. Just given it the same KE earlier on.

7. ### Simon Bridge

15,294
Re: Concept of 'work done' on a smooth slope (high school-level physic

If it is moving at a constant speed - it's kinetic energy does not change.
Work is change in energy.
If the object accelerated, it started out with some kinetic energy, and gained some more kinetic energy. That's the "extra".

8. ### question dude

80
Re: Concept of 'work done' on a smooth slope (high school-level physic

so what you're saying, is that technically there would be K.E, but because it can be so small, it would just be ignored in the calculations?

I don't understand how thats possible in my example. If no more work is done (i.e the 10N force is removed after a while), surely the object would slow down and then fall backwards down the slope?

but if it was on a level resistance-less surface, then yeah that does make sense.

Tell me if I've roughly understood you correctly. So in the previous example in your 1st paragraph where you had the object moving in 1000s, the force applied would be less, but it would be applied over a greater distance. In this second example here, force would be greater, but it would be applied over a lesser distance. So in both examples, if you multiplied force by distance, the energy you get would still be the same.

on the bit in bold, same question as before here, surely you'd still need to have a force (pushing it up the slope) that matches with the object's weight component (which acts down the slope) to keep the object moving upwards at a constant velocity for the rest of the time, otherwise the object would slow down and then fall back downwards

9. ### question dude

80
Re: Concept of 'work done' on a smooth slope (high school-level physic

what I meant is that the object starts off stationary, doesn't it?

I mean, isn't it the force that causes it to move upwards, so before it was applied, the object would've been stationary (or on the verge of falling down the slope), therefore there must've been a change in velocity right at the very start to initiate the movement, right?

unless you take the assumption that the object was somehow already moving upwards at a constant speed before the force applied to it, in which case then yeah I can understand your explanation

10. ### Simon Bridge

15,294
Re: Concept of 'work done' on a smooth slope (high school-level physic

Then it was not moving at a constant speed then was it? Please reread the early answers. I am being entirely consistent.

Recap:
* All the work (moving 5m up the slope under 10N force) goes into potential energy provided the motion was at constant speed. (post #2)
* If there was an acceleration, then some of the work goes into additional kinetic energy as well. (post #4)

It does not matter how this state of affairs came to be - anyway, you didn't say - once the motion is there, this is how the energy and work behaves.

11. ### Philip Wood

1,176
Re: Concept of 'work done' on a smooth slope (high school-level physic

(1) Yes, that's exactly what I'm saying. The KE can be made negligible.

(2) No work to keep it moving at constant speed? I confess to this being ambiguous. You are right to query it. What I mean is this. Suppose the object is going at constant speed up the slope. The work which has to be done on it is work done against gravity and finishes up as the body's grav PE. No work goes into giving the body kinetic energy; not once it's moving at constant speed.

(3) Your first para under the third quote: Yes. Spot-on!

(4) Your last para. You're right, but so am I! I'm saying that you need no excess force, and earlier I wrote "(accelerating)" after excess, to try and explain what I meant. You still need a force (to overcome the component along the slope of the gravitational force), but you don't need any force in excess of this in order to accelerate the object - because it's moving at constant speed!

12. ### question dude

80
Re: Concept of 'work done' on a smooth slope (high school-level physic

Oh I see now, thank you. Yeah I worded my example scenario a bit unclearly, but this is how the questions stated are in my textbook. They don't explicitly say whether or not the object was already moving at that speed just before the force was applied, or whether the object was stationary but because its speed only increased to a very small amount (after force was applied), I could just ignore K.E at the end. Its just expected on you to assume, but fair enough.

13. ### question dude

80
Re: Concept of 'work done' on a smooth slope (high school-level physic

Thanks a lot, it was just a bit of misunderstanding then. Thats a relief.

I just have one more question. I thought of another way of thinking about this scenario, can you tell me if it can be used (if its correct):

- object is stationary on a smooth friction-less slope (technically on the verge of slipping down) and we want to move it 10m up the slope

- a force slightly greater than the weight's down-the-slope component is applied to get it moving at speed 'V' from a stationary start. Lets say this force was 6N

- very soon afterwards (lets say at 0.5m up the slope) when speed has reached 'V', the force applied is reduced so that it is now equal to the object weight's down-the-slope component. Lets this force was 4N

- from the point where force 4N starts to 0.5m before the end of the 10m distance, the object would move at constant velocity 'V', with force 4N staying there to maintain its upward movement for a total distance of 9m (10 - 0.5 - 0.5)

- at 0.5m before the end point, the force is now reduced such that when the object reaches the end of the 10m, its velocity would be zero (as it was at the very start). Lets say this force was 2N

so I would say this is no different from a constant force 4N being applied to a distance 10m up the slope, and for literally all of the 'work done' by the force to be converted into gravitational potential energy

except the force isn't technically constant at 4N throughout the whole 10m, but the overall average = 4N, so the end result is the same as if it was just a constant force all the way through

14. ### Philip Wood

1,176
Re: Concept of 'work done' on a smooth slope (high school-level physic

Agree entirely. Wish I'd thought of that case! No energy 'wasted' as KE overall.

Now I think about it, I remember someone telling me that the tunnels between 'tube' (underground railway) stations in London were curved to be lower half way between the stations than at the stations themselves - in order to 'recover' some kinetic energy from a train as it approaches a station.

Last edited: Feb 16, 2013
15. ### Simon Bridge

15,294
Re: Concept of 'work done' on a smooth slope (high school-level physic

I think we need to see the actual wording in your text book, though, to really see what you are talking about.
You cannot assume anything about the previous state of the object though.
Never assume, you just make an ***ume out of ***ume and assume.  Oh cute!

Note: the same description could refer to an object accelerating down the ramp, or being lowered at a constant speed. After all, if the applied-force were not applied, then the object would be accelerating down the ramp under gravity wouldn't it?

Usually you are given some sort of hint somewhere about what would be a safe assumption. Maybe there is extra information - like the mass of the object and the slope of the ramp(?) maybe the chapter is called "conservation of energy at constant speed"?

Only usually though. Sometimes you have to put some more work in to find out what would be a reasonable guess to make. What's reasonable will be different - you cannot just say "oh I've not been told therefore it's negligible." You need a reason to support that kind of thing.

16. ### question dude

80
Re: Concept of 'work done' on a smooth slope (high school-level physic

well heres an example of a similar question in my textbook:

A luggage trolley of total weight 400N is pushed at steady speed 20m up the slope, by a force 50N acting in the same direction as the object moves in. At the end of this distance, the trolley is 1.5m higher than at the start. Calculate:

a) the work done pushing the trolley up the slope
b) the gain of potential energy of the trolley
c) the energy wasted due to friction

surely I would have to assume it was already moving at that speed just before the pushing force of 50N was applied, because if it was stationary at the very beginning, then there would've been a kinetic energy gain at the end of the 20m distance. The answers to a,b,c shows that the total energy provided by the force to the luggage (work done by force) = potential energy gained + energy wasted due to friction, so thats the calculation they expect you to make which obviously leaves no room for any net kinetic energy gain throughout the whole 20m distance

Last edited: Feb 17, 2013
17. ### Simon Bridge

15,294
Re: Concept of 'work done' on a smooth slope (high school-level physic

Yes and no - the trolly is already moving at a steady speed before the start of the problem because it is already being pushed at the start of the problem - it has already been accelerated to speed by some force or combination of forces we are not told about.

It's like a movie scene - we are just seeing the part of the action that is on screen, who knows how long the trolly was being pushed before we see it? But in the time we see it, it gets pushed 20m at a steady speed. (It may have been pushed at the same force for weeks before we saw it.)