Question

Two blocks, of masses 2kg and 3kg, are connected by a alightstring which passes over a pulley of moment of inertia 0.004kg.m2 and radius 5cm. The coeffiecient of friction for thetable top is 0.30. The blocks are released from rest. Usisng energymethods, one can deduce that after the upper block has moved 0.6m,its speed is: ?

Answer 1

Let $m_1$ and $m_2$ be the masses  

g: acceleration due to gravity,  

I: moment of inertia,  

w: angular speed of the pulley.  

Equating, $PE_i\ of\ m_2\ to\ KE_f$ of the whole system with work done,

$m_2gh = (m_1+m_2)\frac{v^2}{2} + I\frac{w^2}{2} + u m_1 gh$  

v=rw  

Then,

$2(m_2-um_1)gh=v^2[m_1+m_2+\frac{I}{r^2}]$  

When $m_1$=3kg and $m_2$=2kg,

$v=\frac{2(2-0.30*3.0)9.8*0.6} {[3.0+2.0+\frac{0.004}{0.052}]}$

=1.4 m/s  

Answer 2

Answer:

hope it may help you much