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 $$\theta$$ (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 _____.

An object of mass m is placed on a smooth horizontal surface and acted upon by a force F making an angle $$\theta\ $$ with the horizontal, where $$θ=kx$$.

Since the surface is smooth, there is no friction. The vertical component of the force is balanced by the normal reaction, so only the horizontal component contributes to acceleration.

$$F\cos\ \theta\ =ma$$

Now relate acceleration to velocity using:

$$F\cos\left(\ kx\right)\ =m\times\ \frac{\ dv}{dt}$$

$$F\cos\left(\ kx\right)\ =m\times\ \frac{\ dv}{dt}\times\ \ \frac{\ dx}{dx}$$

$$F\cos\left(\ kx\right)\ =m\times\ \frac{\ dv}{dx}\times\ \ \frac{\ dx}{dt}$$

$$F\cos\left(\ kx\right)\ =m\times\ v\times\ \frac{\ dv}{dx}$$

Now separate variables and integrate both sides, taking limits from initial position (where velocity is zero) to any position xxx.

$$F\cos\left(\ kx\ \right)\times\ dx\ \times\ \frac{\ 1}{m} =v\times\ dv$$

$$\int\ F\cos\left(\ kx\ \right)\times\ dx\ \times\ \frac{\ 1}{m}\ =\int\ v\times\ dv$$

$$\ \ \frac{\ F}{k}\sin\ \left(kx\right)\times\ \frac{\ 1}{m}\ =\ \frac{\ v^2}{2}$$

Now express kinetic energy using velocity.

$$\ \ \frac{\ F}{k}\sin\ \left(kx\right)=\ \frac{\ mv^2}{2}=\ K.E$$

Finally, use the relation $$\theta = kx$$ to rewrite the expression in terms of $$\theta\ $$ and compare with the given form.

$$\ \frac{\ F}{k}\sin\theta\ =\frac{n}{k}\sin\theta$$

$$\ \frac{\ 2}{k}\sin\theta\ =\frac{n}{k}\sin\theta$$ (as F = 2)

Thus, the value of n is 2.