In the shown pulley-block system, strings are light. Pulleys are massless and smooth. System is released from rest. In 0.6 second: [Take g=10m/s2]
2T−10g=10a and 20g−T=40a ⇒ 30g=90a ⇒ a=g/3
For (A): W=mgh=(20)(10)[12×g3×(0.6)2]=120J
For (B): T=(10g−10a2)=(10g−−10g/32)=10(2g3)2=(1003)N
⇒W=−T×(h)=−1003×610=−20J
For (C): Wgravity=−mg(h/2)=−10×(10)(620)=−30J
For (D): =+20J