JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Work, Power & Energy

A force

F = α + β x^{2}

acts on an object in the x -direction. The work done by the force is 5 J when the object is displaced by 1 m . If the constant

α = 1 N

then

β

will be

(A) $15 N / m^{2}$
(B) $10 N / m^{2}$
(C) $12 N / m^{2}$ ✓ Correct
(D) $8 N / m^{2}$

Answer: (C)

Solution

Work done by a position-dependent force over a displacement is

W = \int_{0}^{1} F d x = \int_{0}^{1} (α + β x^{2}) d x = [α x + \frac{β x ^{3}}{3}]_{0}^{1} = α + \frac{β}{3}

With

W = 5

J and

α = 1

5 = 1 + \frac{β}{3}

\frac{β}{3} = 4

β = 12 N/m^{2}

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option

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