Question

two blocks of different masses are hanging on two ends of a string passing over a frictionless pulley.the heavier block is twice that of the lighter one. the tension in the string is 60 N. find decrease in PE during the first second after the system is released.

Answer 1

let the mass of lighter and heavier block be m and 2m respectively....
for block m
T-mg=ma................1
for block 2m
2mg-T=2ma...............2
adding 1 and 2
3ma=mg and T=4ma..........1
accleration of both blocks is same so displament after 1 sec for both will be equal
d=ut+at^2/2 (u=0 and t=1s)
d=a/2
lighter block will be displaced upward and heavier block is displaced downward so total potential energy
stored is PE= 2mgd + mg(-d)
=mgd..............2

solving 1 and 2
PE=mga/2=mgT/8m
PE =Tg/8
PE=60*9.8/8
PE=75 J [approx]