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:


 

Option 1 -

At x = x₃, K.E. = 4J.

Option 2 -

At x > x₄, K.E. is constant throughout the region.


Option 3 -

At x < x, K.E. is smallest and the particle is moving at the slowest speed.

Option 4 -

At x = x₂, K.E. is greatest and the particle is moving at the fastest speed.

0 3 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 3


    Detailed Solution:

    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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A
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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

R
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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

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