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`        The potential energy of certain spring when stretched through a distance s is 10 joule.The amount of work that must be done on this spring to stretch it through additional distance s will be`
one year ago

Arun
23371 Points
```							Dear Navya P. E. to stretch through x is =  ½ k* x^2 = 10 to tretch through additional distance x ½ k (4s^2 – s^2 ) = 3/2 * k* s^2 = 3 * 10 = 30 Joules
```
one year ago
Komal
15 Points
```							PE = 1/2kx^2U1= 1/2ks^2 = 10 j ............eq(1) Addition length of spring is =STotal lenght =2sU2=1/2k(2s) ^2W. Done = u2 -u1                 = 1/2k(2s)^2 - 1/2ks^2                = (1/2ks^2)4 - 1/2ks^2  { Here replace the value of 1/2ks^2 =10}                = 10×4 - 10                 =40 - 10                 = 30j                           = 1/2 ×k 4s^2  - 1/2ks^2                  =4(1/2ks^2) -
```
4 months 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