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 \(v(\theta)\) of the bar as a function of the angle \(\theta\) is:

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

Hint:

When force direction varies with position, try converting angle to position using given relation.
Work done = area under force-position graph.
Use energy conservation if force is non-constant or invariable direction.

Answer 1

The Correct Option is B

Step 1: Since the direction of force varies with displacement, we use the work-energy theorem: \[ W = \int F \cos\phi \, dx \] But here, angle \(\theta\) is a function of \(x\): \(\theta = kx \Rightarrow x = \frac{\theta}{k}\) Instead of resolving components, we observe that net work done is: \[ W = \int_0^x F \cos(kx) \, dx \] Or, equivalently: use energy approach — from external force: \[ \frac{1}{2}mv^2 = \int F \cos(kx) dx = \frac{F}{k} \sin\theta \] So, \[ v^2 = \frac{2F \sin\theta}{mk} \Rightarrow v = \sqrt{\frac{2F \sin\theta}{mk}} \]