MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 9
        
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

Answers : (2)

Arun
22821 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^2
U1= 1/2ks^2 = 10 j ............eq(1) 
Addition length of spring is =S
Total lenght =2s
U2=1/2k(2s) ^2
W. 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) - 
2 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details