Question

The diagram below shows a frictionless wheel with a weightless cable wrapped around it and a block attached to the end of the cable. The wheel has a mass of 6.0 kg and the block has a mass of 2.0 kg. The block is released from rest and drops as the cable unwinds. When the block has dropped 1.0 m, its velocity is 2.8 m/s. What is the kinetic energy of the wheel?​

Answer 1

Explanation:

find the acceleration:

v^2 = v0^2 + 2as 

2.8^2 = 2a*1

a = 3.92 m/s^2

use 

torque = (mg - ma)*R = I*α where m is the mass of the block, and α = a/R with R = radius and I = moment of inertia

2(9.81-3.92)*R = I*3.92/R

11.78R^2 = 3.92*I

I = 3.00R^2

the kinetic energy at that point is 

E = 1/2*I*w^2 where w = v/R 

E = 1/2*3R^2*2.8^2/R^2

E = 0.5*3*2.8^2 

E = 11.76 J