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JEE Main 202127 JulyEvening Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2021 · 27 July · Evening Shift · Physics · Work, Power and Energy

Given below is the plot of a potential energy function U ( x ) for...

Given below is the plot of a potential energy function U(x)U ( x ) for a system, in which a particle is in one dimensional motion, while a conservative force F(x)F ( x ) acts on it. Suppose that Emech=8J{E}_{\text{mech}} = 8J, the incorrect statement for this system is :

Given below is the plot of a potential energy function U ( x ) for a system, in which a particle is in one dimensional

Options

  1. Aat x>x4,K.Ex > {x}_{4}, K . E. is constant throughout the region.
  2. Bat x<x1,K.Ex < {x}_{1}, K . E. is smallest and the particle is moving at the slowest speed.
  3. Cat x=x2,K.Ex = {x}_{2}, K . E. is greatest and the particle is moving at the fastest speed.
  4. Dat x=x3,K.E.=4Jx = {x}_{3}, K . E . = 4J

Correct answer

B. at x<x1,K.Ex < {x}_{1}, K . E. is smallest and the particle is moving at the slowest speed.

Step-by-step solution

(A) at x>x4,U=x > {x}_{4}, U = constant =6J= 6J

K=Emech.−U=2J=constantK = {E}_{mech .} - U = 2J = constant

(B) at x<x1,U=x < {x}_{1}, U = constant =8J= 8J

K=Emech.−U=8−8=0JK = {E}_{mech .} - U = 8 - 8 = 0J

Particle is at rest.

(C) At x=x2,U=0⇒Emech.=K=8Jx = {x}_{2}, U = 0 ⇒ {E}_{mech .} = K = 8J
KEKE is greatest, and particle is moving at fastest speed.

(D) At x=x3,U=4Jx = {x}_{3}, U = 4J

U+K=8JU + K = 8J

K=4JK = 4J

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