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Two blocks of different mass are hanging on two ends of a string passing over a friction less pulley.The heavier block has twice the mass as that of the lighter one.The tension in the string is of 60N.The decrease in potential energy during the first second after the system is released is 2k.Then find the value of k

Two blocks of different mass are hanging on two ends of a string passing over a friction less pulley.The heavier block has twice the mass as that of the lighter one.The tension in the string is of 60N.The decrease in potential energy during the first second after the system is released is 2k.Then find the value of k

Grade:11

1 Answers

ANURAG AGARWAL
16 Points
3 years ago
75 is answer.
lets make eq
t-mg=ma
2mg-t=2ma
t=60n
solving, a=10/3m/s sq, m=4.5 kg.
now, use newton 2nd law, u=0
s=1/2 *a*t sq
s=1/2 *10/3 *1
s=5/3=height
poteintal energy=mgh=4.5*10*5/3
=75
 

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