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The potential energy of a long spring when stretched by 2 cm  is U. If the spring is stretched by 8 cm , potential energy stored in it will be:

Option 1 -

16 U

Option 2 -

2 U

Option 3 -

4 U

Option 4 -

8 U

0 2 Views | Posted 4 weeks ago
Asked by Shiksha User

  • 1 Answer

  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    4 weeks ago
    Correct Option - 1


    Detailed Solution:

    Potential energy stored in the spring = 1 2 k x 2

      Now   1 2 k ( 2 ) 2 = U & 1 2 k ( 8 ) 2 = U '   (say)   ⇒ U ' = 64 4 U = 16 U

Similar Questions for you

Two springs with spring constants  k 1 = 1200 N / m and k 2 = 500 N / m  are stretched by the same force. The ratio of potential energy in the springs will be
A
alok kumar singh

Elastic potential energy stored in the spring

U = 1 2 k x 2

= 1 2 F x = F 2 2 k

H e n c e U 1 U 2 = k 2 k 1 F i s s a m e f o r b o t h s p r i n g s

= 500 1200 = 5 12

Given below is the plot of a potential energy function U(x) for a system, in which a particle is in one dimensional motion, while a conservation force F(x) acts on it. Suppose that E_mech = 8J, the incorrect statement for this system is:


 
V
Vishal Baghel

ME = PE + KE = 8J
At x? , PE=4J, so KE = 8-4=4J. (A is correct)
At x>x? , PE=6J, so KE=2J, which is constant. (B is correct)
At x8J. This is not possible. Particle cannot reach here. So it is not a correct statement. (C is incorrect)
At x? , PE=0J, so KE=8J, which is maximum. (D is correct)

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