Rotational Kinetic Energy

Question

I need help with this physics question:

A metal block with mass m = 7.50 kg is connected by a cord wrapped around a 1.25-kg hollow cylinder pulley with a frictionless axle as shown. The surface of the incline has a coefficient of friction μk = 0.30. As the block slides down the inclined plane from rest at an angle of 35 degrees, it pulls the cord and causes the pulley to rotate. If the inclined plane has a length of 2.25 m, calculate the speed v of the block before it reaches the bottom. The pulley has a moment of inertia of a solid disk, I = (mr^2)/2 , and a radius r = 0.25 m.

Answer 1

Dynamics approach:

M = 7.50kg, m = 1.25kg, g = 9.81m/s²

Forces on the box:

Wsin(35°) - T - Wμcon(35°) = M*a

Forces on the pulley:

T*r = I*α ----> T*r = mr² * α/2 ----> T= mrα/2 ----> α*r = a ----> T= ma/2

Wsin(35°) - ma/2 - Wμcon(35°) = M*a

Mgsin(35°) - Mgμcon(35°) = M*a+ ma/2

a = [Mgsin(35°) - Mgμcon(35°)]/[M+(m/2)]

a = 2.97 m/s²

v² = (vo)² + 2*a*d vo = 0, d=2.25m

v² = 2*2.97*2.25

v = 3.65 m/s