Same Acceleration, Same Time? Exploring Work and Power in Inclined Motion

[xxWhiteWalkerxx]

Homework Statement

Paul and Mark ran up an incline. Paul is twice as massive as Mark. Mark reached the top of the incline in half the time. Who exerted more work? Who was more powerful?

2. Homework Equations
Power = work/time; work = force x distance; force = mass x acceleration

3. The Attempt at a Solution
teacher's answer: Same power but Paul exerted more work.

This was my son's hw. My confusion is with regard to their respective accelerations. The teacher assumed the same acceleration (but didn't state it as a given). My confusion -- shouldn't objects of the same acceleration cover the same distance at the same time? Isn't this the same as free falling bodies? So Paul and Mark should have reached the top of the incline at the same time if they have the same acceleration. Paul was more powerful and worked harder to get there at the same time. Am I correct?

Help me out. Thanks in advance :)

 

[axmls]

I agree the question is a little vaguely worded, but I believe it means to say that the runners are running with a constant velocity.

 

[256bits]

force = mass x acceleration

Yes , that can be true.
It also can be true that there can be a force in moving an object without an acceleration which means at a constant velocity, such as pushing a box across the floor - one has to provide a force against the friction force to keep the box moving.

In the problem, the easiest and most intuitive method to solve is by assuming both Paul and Mark are running at a constant velocity, but with Mark at a speed twice as fast as Paul's. That does take care of the fact stated in the problem that Mark reached the top in half the time. In this case Paul and Mark would have the same acceleration = 0.

 

[xxWhiteWalkerxx]

I understand it now (kind of - intuitively speaking) but I need to read up. I thought i was done after college (apparently not if one has kids - sigh). Thank you guys for the help. Much appreciated :)

 

[RUber]

I agree that some information is missing from the question. I would tend to read this as both runners starting from zero and accelerating up the hill, but in the context of the question, that probably isn't what the teacher is asking for, since no details are given.

If acceleration is zero, then the only force exerted is that which is required to overcome gravity. This is the same for both, since they are on the same incline.
If Mark's mass is M and Paul's mass is 2M and the acceleration required to overcome gravity is A, then Paul's force = 2MA and Mark's force is MA.
Mark's time is T/2 and Paul's time = T, so Power is the same. Distance is the same for both, so Paul's work is greater (twice Mark's).

If you were to assume constant acceleration from a full stop, then you would use D = AT^2/2 to find relative accelerations. This would show you that for Mark to make it up the hill in half the time, his acceleration would need to be four times Paul's. In this circumstance, Mark's force = 4MA and Paul's force = 2MA. So Mark would have done twice the work in half the time, giving him 4 times the power of Paul .

 

[xxWhiteWalkerxx]

^this one cleared it for me. I was about to ask how to show it using the equations i posted above when the alert showed a new post :) Thanks.

 

[haruspex]

If you were to assume constant acceleration from a full stop, then you would use D = AT^2/2 to find relative accelerations. This would show you that for Mark to make it up the hill in half the time, his acceleration would need to be four times Paul's. In this circumstance, Mark's force = 4MA and Paul's force = 2MA. So Mark would have done twice the work in half the time, giving him 4 times the power of Paul .

You get this result because you don't allow for them slowing down. If they have residual KE at the top then you cannot tell how much work they have done.
The most appropriate assumption is that they coast to a stop at the top, as you do, and that the question is asking for the average power, $\frac{\Delta E}{\Delta t}$.

 

[RUber]

Thanks @haruspex for clearing that up. That would also support the assumption of constant velocity leading to the teacher's answer.