JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 7)
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 :
_27th_July_Evening_Shift_en_7_1.png)
[ where K.E. = kinetic energy ]
[ where K.E. = kinetic energy ]
at x > x4 K.E. is constant throughout the region.
at x < x1, K.E. is smallest and the particle is moving at the slowest speed.
at x = x2, K.E. is greatest and the particle is moving at the fastest speed.
at x = x3, K.E. = 4 J.
Explanation
Emech. = 8J
(A) at x > x4
U = constant = 6J
K = Emech. $$-$$ U = 2J = constant
(B) at x < x1
U = constant = 8J
K = Emech $$-$$ U = 8 $$-$$ 8 - 0 J
Particle is at rest.
(C) At x = x2,
U = 0 $$\Rightarrow$$ Emech. = K = 8 J
KE is greatest, and particle is moving at fastest speed.
(D) At x = x3
U = 4J
U + K = 8J
K = 4J
(A) at x > x4
U = constant = 6J
K = Emech. $$-$$ U = 2J = constant
(B) at x < x1
U = constant = 8J
K = Emech $$-$$ U = 8 $$-$$ 8 - 0 J
Particle is at rest.
(C) At x = x2,
U = 0 $$\Rightarrow$$ Emech. = K = 8 J
KE is greatest, and particle is moving at fastest speed.
(D) At x = x3
U = 4J
U + K = 8J
K = 4J
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