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 $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta=k x$, 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 $E=\frac{n}{k} \sin \theta$ The valte of $n$ is

Answer 1

Correct Answer: 2

To find the value of n, we will use the relationship between force, displacement, and kinetic energy. Given that the angle θ varies as θ=kx:

1. Consider the horizontal component of the force F: F_h=F·cos(θ).
2. The work done W by F_h over distance x is:
   W=∫F_hdx=2∫cos(kx)dx.
3. The integral gives:
   W=2·[sin(kx)/k]+C, where C is the constant of integration.
4. Since the object starts from rest, initial kinetic energy K.E. is zero, making C=0.
5. The kinetic energy E is given as:
   E=W=2·sin(kx)/k.
6. By comparing this with E=n·sin(θ)/k, we find n=2.

Thus, the value of n is 2, which lies within the expected range [2,2].