JEE Main 2025 Jan 24 Shift 1, Physics Q19: Work Energy Theorem And Its Applications

JEE Main2025Jan 24, Shift 1Physics

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 $^{2}$
D
$8$ N/m $^{2}$

Solution

Work done by a position-dependent force in one dimension:

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

Set $W = 5$ J and $α = 1$ N:

$1 + \frac{β}{3} = 5 \Rightarrow β = 12 N/m^{2}$

$∴$ Option (3).

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JEE Main 2025 Jan 24 Shift 1, Physics Q19: Work Energy Theorem And Its Applications

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