# Conservation of Energy in a Pulley System

1. Dec 5, 2015

### StinkLog

1. The problem statement, all variables and given/known data
Two masses are connected by a light string passing over a light, frictionless pulley, as shown in the figure below. The mass m1 (which is greater than m2) is released from rest. Use the isolated system model to answer the following.

In terms of m1, m2, and h, determine the speed of m2 just as m1 hits the ground (Use m_1 for m1, m_2 for m2, g, and h as appropriate.)
2. Relevant equations
Ug=mgh
KE=.5mv^2
3. The attempt at a solution
Since energy is conserved, I figured that the potential energy of m1 would equal the kinetic energy of m2.
I used the equation:
Ug=KE
m_1*g*h=.5*m_2*v^2
sqrt((2*m_1*g*h)/m_2)=v
This answer is incorrect so I was thinking that perhaps that m2 would raise to the same height h as m2 dropped and that m2 would have some kinetic and some potential energy.
I used this equation:
Ug=Ug+KE
m_1*g*h=m_2*g*h+.5*m_2*v^2
sqrt((2*m_1*g*h-2*m_2*g*h)/m_2)=v
This answer was also incorrect so I'm not sure where to go from here.

2. Dec 5, 2015

### Staff: Mentor

You were correct to take into account the changes in potential energy of both masses. But the kinetic energies of both masses also change.

Chet

3. Dec 5, 2015

### StinkLog

ohhh so the energy at the beginning would be just potential energy and at the end it would be the potential energy of m2 plus the kinetic of m1 and of m2
Ug=Ug+KE+KE
m_1*g*h=m_2*g*h+.5*m_2*v^2+.5*m_1*v^2
sqrt(2*(m_1*g*h-m_2*g*h)/(m_2+m_1)=v