A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg . Taking g=10 m / ^ 2 , find the work done (in joules) by the string on the block

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 8 J

Given m=0.36 kg, M=0.72 kg . The figure shows the forces on m and M . When the system is released, let the acceleration be a . Then T-m g=m aM g-T=M a a= ((M-m) g)/(M+m)= g / 3 and T=4 mg / 3 For block m: u=0, a=g / 3, t=1, s=? s=u t+(1)/(2) a t^ 2 =0+ (1)/(2) × (g)/(3) ×1^ 2 =g / 6 Work done by the string on m is T s =T s=4 (m g)/(3) × (g)/(6)=(4 × 0.36 × 10 × 10)/(3 × 6)=8 J

A light inextensible string that goes over a smooth fixed pulley as...

Options

Correct answer

Step-by-step solution

More from Work, Power and Energy