JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Work, Power & Energy

A particle experiences a variable force

F = (4 x \hat{i} + 3 y^{2} \hat{j})

in a horizontal

x - y

plane. Assume distance in meters and force is newton. If the particle moves from point

(1, 2)

to point

(2, 3)

in the

x - y

plane, the Kinetic Energy changes by

(A)50.0 J
(B)12.5 J
(C)25.0 J✓ Correct
(D)0 J

Answer: (C)

Solution

The work done equals the change in kinetic energy. For the given force:

W = \int F \cdot d r = \int_{1}^{2} 4 x d x + \int_{2}^{3} 3 y^{2} d y

W = [2 x^{2}]_{1}^{2} + [y^{3}]_{2}^{3}

W = (8 - 2) + (27 - 8) = 6 + 19 = 25 J

Therefore

Δ K E = 25 J

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option

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