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 d

Question

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 = k x , 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 k sin θ . The value of n is ______.

TestHub · Accepted Answer

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

Using Newton's law of motion

Along horizontal direction, $F \cos θ = ma$

$\Rightarrow 2 \cos (k x) = m v \frac{d v}{d x}$

Integrating the above relation, we have

$\int_{0}^{v} v d v = 2 \int_{0}^{x} \cos (k x) d x$

$\Rightarrow \frac{m v^{2}}{2} = \frac{2}{k} \sin k x$

Thus, kinetic energy of the object will be $K . E . = \frac{2}{k} \sin θ$

Thus, the value of $n = 2$ .

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