# Law of Conservation of Mechanical Energy Exercise

1. Feb 8, 2016

### Kernul

1. The problem statement, all variables and given/known data
A mass $m_1 = 5.0 kg$ is hanging from the end of a thread, of negligible mass, that slides on a pulley, of negligible mass too and without friction. At the other end of the thread, at the same height of $m_1$, there is another hanging mass $m_2 = 3.5 kg$. Using the Law of Conservation of Mechanical Energy, determine the velocity of $m_1$ when it goes down, starting still, of a height of $2.5 m$.

2. Relevant equations
$E = K + U$
$K_f + U_f = K_0 + U_0 \Rightarrow E = const$

3. The attempt at a solution
So, since we have two masses that move at the same time but have different masses, we can start writing these:
$K_0 = \frac{1}{2} * m_1 * v_0^2 + \frac{1}{2} * m_2 * v_0^2$
But since we know it starts still, $v_0 = 0$ and so $K_0 = 0$.
$U_0 = 0$ because it hasn't moved yet.
$K_f = \frac{1}{2} * m_1 * v_f^2 + \frac{1}{2} * m_2 * v_f^2$
with $v_f$ being the velocity we are searching for.
$U_f = m_2 * g * h - m_1 * g * h$
So, using the second relevant equation we have:
$K_f = - U_f$
$\frac{1}{2} * m_1 * v_f^2 + \frac{1}{2} * m_2 * v_f^2 = m_1 * g * h - m_2 * g * h$
$\frac{v_f^2}{2}(m_1 + m_2) = g * h(m_1 - m_2)$
$v_f =\sqrt{ \frac{2 * g * h(m_1 - m_2)}{(m_1 + m_2)}} = 3 \frac{m}{s}$
Is this method I used to find the velocity with the law good? Or I could have done better and faster?

2. Feb 8, 2016

### BvU

This is about it, can't do much better than $\Delta K + \Delta U = 0$

3. Feb 8, 2016

### Kernul

Okay! Thank you for replying! :D

4. Feb 8, 2016

### BvU

You're welcome.