Q67 / 90 · JEE Main 2021 (27 Jul Shift 2) · Physics

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 conservative force F(x) acts on it. Suppose that Emech =8 J, the incorrect statement for this system is :

Figure

  • A
    at x>x4, K.E. is constant throughout the region.
  • Correct answer
    B
    at x<x1, K.E. is smallest and the particle is moving at the slowest speed.
  • C
    at x=x2, K.E. is greatest and the particle is moving at the fastest speed.
  • D
    at x=x3, K.E.=4 J

Solution

(A) at x>x4,  U= constant =6 J

K=Emech. -U=2 J= constant

(B) at x<x1,  U= constant =8 J

K=Emech. -U=8-8=0 J

Particle is at rest.

(C) At x=x2,U=0⇒Emech. =K=8 J
KE is greatest, and particle is moving at fastest speed.

(D) At x=x3,  U=4 J

U+K=8 J

K=4 J

Hence, the correct answer is option: 'B'.

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