Question

A block of mass \( m \) with an initial kinetic energy \( E \) moves up an inclined plane of inclination \( \theta \). If \( \mu \) is the coefficient of friction between the plane and the body, the work done against friction before coming to rest is:

• A. \( \mu E \cos \theta \)
• B. \( \frac{\mu E \cos \theta}{\sin \theta - \mu \cos \theta} \)
• C. \( \frac{E \mu \cos \theta}{\cos \theta + \sin \theta} \)
• D. \( \frac{\mu E \cos \theta}{\sin \theta + \mu \cos \theta} \)

Hint:

Use energy conservation to find work done against friction. - Frictional force always acts opposite to motion on an inclined plane.

Answer 1

The Correct Option is D

Step 1: Work done against friction
The force of friction acting on the block is: \[ F_f = \mu mg \cos \theta. \] The work done against friction is given by: \[ W = F_f \times d. \] Step 2: Solve for \( d \)
Using energy conservation: \[ E = mgd (\sin \theta + \mu \cos \theta). \] \[ d = \frac{E}{mg (\sin \theta + \mu \cos \theta)}. \] Step 3: Compute work against friction
\[ W = \mu mg \cos \theta \times \frac{E}{mg (\sin \theta + \mu \cos \theta)}. \] \[ W = \frac{\mu E \cos \theta}{\sin \theta + \mu \cos \theta}. \] Thus, the correct answer is \( \boxed{\frac{\mu E \cos \theta}{\sin \theta + \mu \cos \theta}} \).