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 s

Step-by-step solution

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 ⇒ 2 cos ( k x ) = m v d v d x Integrating the above relation, we have ∫ 0 v v d v = 2 ∫ 0 x cos ( k x ) d x ⇒ m v 2 2 = 2 k sin k x Thus, kinetic energy of the object will be K . E . = 2 k sin θ Thus, the value of n = 2 .