An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force F = 2N. In the process of its linear motion, the angle θ (as shown in figure) between the direction of force and horizontal varies as θ = kx, where k is a constant and x is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be E = n/ksin θ. The value of n is ___________. Numerical answer type. Work it out, then open the answer.

Answer: 2. Only the horizontal component does work on the smooth floor: dW = Fcos θ dx with θ = kx. E = ∫₀x 2cos (kx) dx = 2/ksin (kx) = 2/ksin θ Comparing with E = n/ksin θ: n = 2.

An object of mass 'm' initially at rest on a smooth horizontal plane… | JEE Main 2023 (25 Jan, Shift 1)

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Work by a Variable Force

An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force F = 2N. In the process of its linear motion, the angle $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta = kx$ , where k is a constant and $x$ is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be $E = \frac{n}{k}\sin\theta$ . The value of n is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Only the horizontal component does work on the smooth floor: $dW = F\cos\theta\,dx$ with $\theta = kx$ .
$E = \int_0^x 2\cos(kx)\,dx = \frac{2}{k}\sin(kx) = \frac{2}{k}\sin\theta$
Comparing with $E = \frac{n}{k}\sin\theta$ : $n = 2$ .

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