Question

A bar of mass \(m\) resting on a smooth horizontal plane starts moving due to a constant force \(F\). In the process of its rectilinear motion, the angle \(\theta\) between the direction of this force and the horizontal varies as \(\theta = kx\), where \(k\) is a constant and \(x\) is the distance traversed by the bar from its initial position. The velocity \(v\) of the bar as a function of the angle \(\theta\) is

• A. \(v=\sqrt{\dfrac{2F\sin\theta}{mk}}\)
• B. \(v=\sqrt{\dfrac{2F}{mk\sin\theta}}\)
• C. \(v=\dfrac{2F\sin\theta}{mk}\)
• D. \(v=\dfrac{2F}{mk\sin\theta}\)

Hint:

When force acts at a variable angle with displacement, use the work-energy theorem: \[ dW=F\cos\theta\,dx. \] Then use the given relation between angle and displacement to integrate properly.

Answer 1

The Correct Option is A

Step 1: Apply the work-energy theorem.
On a smooth horizontal plane, only the horizontal component of force \( F \) does work. With \( \theta = kx \), the horizontal component at position x is \( F\cos(kx) \). Work done to reach displacement \( x = \theta/k \): \[ W = \int_0^{\theta/k} F\cos(kx)\,dx = \frac{F}{k}\sin\theta \]

Step 2: Equate work to kinetic energy.
\[ \frac{1}{2}mv^2 = \frac{F}{k}\sin\theta \Rightarrow v = \sqrt{\frac{2F\sin\theta}{mk}} \] \[ \boxed{v = \sqrt{\dfrac{2F\sin\theta}{mk}}} \]