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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

```
2 months ago

ANURAG AGARWAL
16 Points
```							75 is answer.lets make eqt-mg=ma2mg-t=2mat=60nsolving, a=10/3m/s sq, m=4.5 kg.now, use newton 2nd law, u=0s=1/2 *a*t sqs=1/2 *10/3 *1s=5/3=heightpoteintal energy=mgh=4.5*10*5/3=75
```
one month ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions