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An object of mass m initially at rest on a smooth horizontal plane starts moving under the action of force F=2 N. 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=nksin θ. The value of n is ______.

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Solution:
1494 Upvotes Verified Answer
The correct answer is: 2

As seen in the free body diagram of the object, normal force N, weight mg and force in parallel and perpendicular directions are acting.

Using Newton's law of motion

Along horizontal direction, Fcosθ=ma

2cos(kx)=mvdvdx

Integrating the above relation, we have

0vvdv=20xcos(kx)dx

mv22=2ksinkx

Thus, kinetic energy of the object will be K.E.=2ksinθ

Thus, the value of n=2.

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