Problem 3. Two blocks are connected by a string running over a cylindrical pulle

Question

Problem 3. Two blocks are connected by a string running over a cylindrical pulley. The blocks have mass of 3.0 kg and 5.7 kg and the pulley has a radius of 0.12 m and mass of 10.3 kg. If the 5.7 kg mass is placed on a tabletop, whereas the 3.0 kg is hanging off on the side, and if the hanging mass descends down by 1.5 m with the system staring from rest, find the speed of each block. Neglect any friction. Problem 3. Two blocks are connected by a string running over a cylindrical pulley. The blocks have mass of 3.0 kg and 5.7 kg and the pulley has a radius of 0.12 m and mass of 10.3 kg. If the 5.7 kg mass is placed on a tabletop, whereas the 3.0 kg is hanging off on the side, and if the hanging mass descends down by 1.5 m with the system staring from rest, find the speed of each block. Neglect any friction.

Explanation / Answer

let

m1 = 3 kg

m2 = 5.7 kg

M = 10.3 kg

r = 0.12 m

let v is the speed of the blocks and w is the angular speed of the pulley

Apply conservation of energy

kinetic energy gained = loss of potentail energy

0.5*m1*v^2 + 0.5*m2*v^2 + 0.5*I*w^2 = m1*g*h

0.5*m1*v^2 + 0.5*m2*v^2 + 0.5*0.5*M*r^2*w^2 = m1*g*h

0.5*m1*v^2 + 0.5*m2*v^2 + 0.5*0.5*M*(r*w)^2 = m1*g*h

0.5*m1*v^2 + 0.5*m2*v^2 + 0.25*M*v^2 = m1*g*h

v^2*(0.5*m1 + 0.5*m2 + 0.25*M) = m1*g*h

v = sqrt(m1*g*h/(0.5*m1 + 0.5*m2 + 0.25*M))

= sqrt(3*9.8*1.5/(0.5*3 + 0.5*5.7 + 0.25*10.3))

= 2.52 m/s <<<<<--------Answer

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