Work done and Power

1. Dec 22, 2012

pratyushag

1. The problem statement, all variables and given/known data
Ram and Shyam both climb a hill.
They have the same mass.
Ram wins the race and reaches in 5 Minutes but shyam reaches in 7 mins.Choose the correct Statement:

A)Ram and shyam do the same work but exert different power
B)Ram and shyam do different work and exert different power
C)Ram and shyam do different work but same different power

2. Relevant equations
F=ma Work=F*S power=Work/Displacement

3. The attempt at a solution
Work=F*S
=ma*S
both have same mass and displacement.... however the acceleration maybe equal or unequal
hence we cannot say if work is equal and consequently we cannot say if power is equal or unequal..

But this seems to deviate from the options provided in the question!

2. Dec 22, 2012

chiro

Hey pratyushag.

For this one the first thing that comes to mind is that the potential energy used to take both people up the mountain should be the same if the path is the same.

The only thing is though that the rate is different which will affect the work done.

3. Dec 22, 2012

pratyushag

I understand what you are saying at first the same concept of mgh came to my mind but then i started thinking F=ma should also work here... but why isnt that giving an answer?

4. Dec 22, 2012

chiro

Think of integrating force with respect to distance (which is work). If the time is smaller for the same path, what does this imply about the force? (given that the distance is the same)?

5. Dec 22, 2012

haruspex

Using force makes it unnecessarily complicated for this, but here's what would happen.
Briefly, Ram would accelerate more than Shyam. Once up to speed, neither is accelerating. The work they are doing is in overcoming gravity, which is the same for both. The initial little extra work that Ram did has gone into giving him extra kinetic energy.
As they near the top, Ram decelerates more than Shyam does, so does a little less work near the end. This exactly compensates the extra work he did at the beginning.