Question

Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has as moment of inertia `1.6xx10^-4kg-m^2` and a radius 2.0 cm. Find a. the kinetic energy of the system as the 400 block falls thrug 50 cm b. the speed of the blcok at this instant.

Answer 1

Correct Answer - A::B::D
According to the question
`04g-T_1=00.4a`……..1
`T_2-0.2g=0.2a`……2
`(T_1-T_2)r=(la)/r`……..3
From eq. 1, 2, and 3
`rarr a=((0.4-0.2)g)/((0.4+0.2+1.6/0.4))=g/5` ltb. Therefoee b. `V=sqrt(2gh)=sqrt(2xgxx1/20`
`rarr sqrt((g/5))=sqrt((9.8/5))=1.4m/s`
a. total kinetic energy of the system
`=1/2m_1V^2+1/2m_2V^2+1/2 15^`
` =(1/2xx0.4xx1.4^2)+(1/2xx0.2xx1.4^2)+ {1/2xx(1.6/4xx1.4^2)}`
`=(0.2+0.1+0.2)(1.4)^2`
`=0.54x1.96=0.98Joule`