# Homework Help: Work, energy and signs

1. Jun 22, 2015

### Lord Anoobis

μ1. The problem statement, all variables and given/known data
A simple diagram shows a block of mass M on a horizontal table with a light cord running from it to the right over a pulley at the edge of the table. At the other end of the cord hangs a block of mass m, which is a height h above the ground. The problem is to derive an expression for the speed of the block m just before it hits the floor if the coefficient of friction for the block on the table is μk.
2. Relevant equations

3. The attempt at a solution
Firstly, let me say that my second attempt at this problem led to the correct answer. However, this is of little consolation, because the reason behind the error in the first effort and its subsequent rectification is not clear.

I used the tension in the cord $T = \frac{mMg(μ_k + 1)}{m + M}$ to determine the work done by that force, the friction the already incorporated. This work, $W_T$ is a negative quantity, so

$W_T = -\frac{mMgh(μ_k + 1)}{m + M}$

Then followed $K + W_T = U_g$ leading to

$\frac{1}{2}mv^2 - \frac{mMgh(μ_k + 1)}{m + M} = mgh$

And some manipulation leads to the incorrect expression. Changing the negative sign to positive in the above equation gives the correct answer. What is the error here?

2. Jun 22, 2015

### jbriggs444

This being an expression for the work done by tension on the hanging mass m, correct?

Can you flesh out how you obtained this equation?

Ordinarily, one would start with Total Initial Energy + Work Done on System = Total Final Energy.

3. Jun 22, 2015

### Lord Anoobis

Yes, that expression is the work done by tension. For the other, I added the kinetic energy of the block at ground level to the work done by tension and set it equal to the initial gravitational potential energy.

4. Jun 22, 2015

### jbriggs444

So that would be:

Final Total Energy + Work Done on System = Initial Total Energy.

Can you see the difference between that and the starting point that I suggested?

5. Jun 22, 2015

### Lord Anoobis

Sorry for the delay, I rushed back as quickly as possible. What I see now is that with $K + W_T = U_g$, I basically accounted for work done on the system twice, correct?

6. Jun 22, 2015

### jbriggs444

Lots of ways to look at it.

Work went on the wrong side of the equal sign. Or...
The work had the wrong sign. Or...
The work was evaluated on the forward path but the equation applied to the reversed path. Or...
The work was added when it should have been subtracted.

But yes, the net effect is to have an error equal to twice the magnitude of the work.

7. Jun 22, 2015

### Lord Anoobis

A valuable lesson learned. Thanks a bunch.