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 Emech =8J, the incorrect statement for this system is:
[where K.E. = kinetic energy]
(A)at x=x3, K.E. =4J.
(B)at x>x4, K.E. is constant throughout the region.
(C)at x<x1, K.E. is smallest and the particle is moving at the slowest speed.✓ Correct
(D)at x=x2, K.E. is greatest and the particle is moving at the fastest speed.
Answer: (C)
Solution
The kinetic energy at any point is K=Emech−U(x)=8−U(x) joule. At x=x3 the curve reads U=4 J, so K=4 J. At x=x2 the potential energy is at its minimum, U=0, so K=8 J is the largest and the speed there is greatest. For x>x4 the curve is flat at U=6 J, so K=2 J and stays constant through that region. For x<x1 the curve is flat at U=8 J, which equals the total energy, so K=0 and the particle is at rest rather than creeping along. The statement that for x<x1 the kinetic energy is smallest and the particle moves at the slowest speed is therefore the incorrect one.
JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option
