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 a moment of inertia 1·6×10-4 kg-m2 and a radius 2⋅0 cm, Find (a) the kinetic energy of the system as the 400 g block falls through 50 cm, (b) the speed of the blocks at this instant.
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The kinetic energy of the system is 0.98 J and the speed of the blocks is 1.4 m/s
Explanation:
According to the question
04g−T1=00.4a……..1
T2−0.2g=0.2a……2
(T1−T2)r=lar……..3
From eq. 1, 2, and 3
a. total kinetic energy of the system
= 1/2m1V2+1/2m2V2+12/15
(1/2×0.4×1.42)+(1/2×0.2×1.42)+{1/2×(1.6/4×1.4^2)}
=(0.2+0.1+0.2)(1.4)2
=0.54 x 1.96=0.98J
→a = (0.4−0.2)g/(0.4+0.2+1.60.4)=g/5 ltb.
Therefore b. V=2gh−−−√=2 x g×1/20
→(g/5) = √(9.8/5) = 1.4 m/s
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