1. Aug 19, 2014

### physics801

I'm studying for my college's physics placement and I'm frustrated because I don't get the solution to problem 1d.

1. The problem statement, all variables and given/known data

The problem is here: http://ocw.edu.ht/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/exams/exam_2.pdf [Broken]

Solution manual is here: http://ocw.edu.ht/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/exams/quiz2_sol.pdf [Broken]

3. The attempt at a solution

I drew a mspaint diagram of what I have: http://cl.ly/image/3b1o082l1k3a ($x_{max}$ is defined in 1a, which is in the soln manual)

Basically since gravity is opposing the direction of travel (up the ramp), I think the solution to 1d(i) should be $work=-mg(h-h_{max})=-mgsin\theta(x-x_{max}).$ However the solution manual has the same thing except $x_{max}-x$ in the parentheses. Why is this? Shouldn't it be $final - initial$, which means $x-x_{max}$ in the parentheses?

edit: I also have the same question for 1d(ii) and (iii) but hopefully it will be answered by explaining (i)

Last edited by a moderator: May 6, 2017
2. Aug 20, 2014

### haruspex

You have to decide which way is positive and stick to it.
If the force due to gravity is being taken as negative along the slope then the positive direction is up the slope. But x is measured down the slope, so it's (-x) - (-xmax)
It is always a good idea to sanity-check the sign of a result. It is clear that the work done by gravity here is negative.

3. Aug 20, 2014

### /AH

Look at the direction of x, what does x mean in terms of the spring?

4. Aug 20, 2014

### physics801

Oh whoops, I completely missed that. Thanks for the help!