The two blocks are 3kgsmaller and 5kg bigger the pulleys mas

The two blocks are 3kg(smaller) and 5kg( bigger), the pulleys mass is negligible. The blocks are initially at rest. Use energy conservation to find the velocity v of the blocks when m2 falls a distance of h=2m.

b) What is the velocity if we know the pulleys mass is 2kg?

Solution

Initial Potential Energy = (m1+m2)*g*h
Initial Kinetic Energy = 0
Final Potential Energy = m1*g*2h
Final Potemtial Energy = 1/2 * m1*v^2

Using Energy Conservation
(m1+m2)*g*h =  m1*g*2h + 1/2 * (m1+m2)*v^2
(3 + 5)*9.8 *2 = 3*9.8*2*2 + 1/2 * (3 + 5) * v^2
v = 3.13 m/s

Velocity of the blocks, v = 3.13 m/s

(b)
Rotational Kinetic Energy of Pulley , K.E = I/2 * Iw^2
Where,
I = 1/2 * M*r^2
w = v/r
K.E = 1/2 * M*r^2 * v^2/r^2
K.E = 1/2 * M*v^2

(m1+m2)*g*h =  m1*g*2h +  1/2 * (m1+m2)*v^2 + 1/2 * M*v^2
(3 + 5)*9.8 *2 = 3*9.8*2*2 + 1/2 * (3 + 5) * v^2 + 1/2 * 2 * v^2
v = 2.8 m/s
Velocity with pulley mass, v = 2.8 m/s