Question

A block of mass M_2 = 6.0 kg is initially at rest on a level table. A string of negligible mass is connected to M_2, runs over a friction less pulley, of 2.0 kg mass and 0.1m radius and is attached to a hanging mass M_1 =5.0 kg 3m above the ground as shown in the figure A. The system was released and the velocity of M_1 was 2.7 m/s when it was 2.0 m above the ground as shown in figure B. Calculate: 

The initial total energy of the system (at A). 

The translational kinetic energy of the two masses at B. 

The rotational kinetic energy of the pulley at B. 

The system total energy at B.(add the potential energy of M_1) 

The work done by friction between M_2 and the table.

Answer 1

M2 = 6 Kg

M1 = 5 Kg

m = 2 Kg

r = 0.1 m

vi = 0 ,

v = 2.7 m/s

h = 3 - 2 = 1 m

total initial energy of system = M1 * g * h

total initial energy of system = 5 * 9.8 * 3

total initial energy of system = 147 J

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at the point b

translational energy of masses = 0.5 * M1 * v^2 + 0.5 * M2 * v^2

translational energy of masses = 0.5 * (6 + 5) * 2.7^2

translational energy of masses = 40.1 J

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rotational energy of pulley = 0.5 * I * (v/r)^2

rotational energy of pulley = 0.5 * 0.5 * 2 * 0.1^2 * (2.7/.1)^2

rotational energy of pulley = 3.645 J

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system total energy at B = 40.1 + 3.645 + 5 * 9.8 * 2

system total energy at B = 141.75 J

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work done by frictional force = 141.75 - 147

work done by frictional force = -5.25 J