Work-Energy Principle: Pulley System

[matineesuxxx]

Homework Statement

(see attatchment)

We know masses M and m, and I am trying to describe the downward speed of m in terms of the height from the ground. There is no friction

Homework Equations

$\sum \text{F} = ma$
$W_{\text{total}} = \Delta K = \int \text{F} \cdot d\text{s}$

The Attempt at a Solution

I first need to note that the distance moved by $M$ is half of that moved by $m$ and thus the acceleration of $M$ is half that of $m$, or $a_M = a_m/2$.

I first solved the problem using only Newtons laws, so from the FBD I solved for the acceleration of $m$:

$$a_m = v\frac{dv}{dh} = \frac{mg}{M/4 + m} \implies v = \sqrt{\frac{2mgh}{M/4 + m}}$$

Now if I use the Work-Energy principle, I get that $W_{\text{total}} = W_{\text{tension}_x} + W_{\text{tension}_y} + W_g$.

This is where I think I am not understanding something:

Work done by tension on $M$ is equal to the work done by tension on $m$ even though force of tension on $M$ is twice that of the force of tension on $m$ because it travels half the distance, and since tension acting on $m$ is acting opposite the direction of motion, it is negative, so $W_{\text{tension}_x} = -W_{\text{tension}_y}$, so

$$W_{\text{total}} = W_g = mgh = \frac{1}{2}(m + M)v^2 = \Delta K $$
so

$v = \sqrt{\frac{2mgh}{m+M}}$

I am not sure what I am missing that gives me $m+M$ instead of $m + M/4$ in the denominator. I would appreciate some help!

Thanks

[EDIT] The answer in the back of the book is the same as the answer I got using Newtons laws

 

[Tanya Sharma]

$$W_{\text{total}} = W_g = mgh = \frac{1}{2}(m + M)v^2 = \Delta K $$

The masses are moving with different speeds .

 

[matineesuxxx]

The masses are moving with different speeds .

Hah, wow. Total brain fart there. Thank you very much!

 

[Tanya Sharma]

Even the answer you are getting by force approach is incorrect .

 

[matineesuxxx]

Even the answer you are getting by force approach is incorrect .

How is that? It's the same as in the book as well

 

[Tanya Sharma]

Sorry ... its correct .

 

[matineesuxxx]

Thanks for your help.

 

[Tanya Sharma]

You are welcome :)