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

A, B, D

Explanation

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`