Question

A small block of mass \(m\) is kept on a rough inclined surface of inclination \(\theta\) fixed in an elevator. The elevator rises with a uniform velocity \(v\), and the block does not slide on the wedge. The work done by the force of friction on the block in time \(t\) will be:

• A. \(0\)
• B. \(mgvt\cos^2\theta\)
• C. \(mgvt\sin^2\theta\)
• D. \(mgvt\sin 2\theta\)

Hint:

Friction balances the along-incline gravity component, \(f = mg\sin\theta\). Multiply by the vertical rise \(vt\) and the vertical component \(\sin\theta\) of the friction direction.

Answer 1

The Correct Option is C

Step 1: The elevator rises with uniform velocity, so there is no acceleration. The block is in equilibrium. The forces on the block are its weight \(mg\) (down), the normal force from the incline, and static friction \(f\) along the incline.
Step 2: Balancing forces along the incline (the block does not slide), friction must balance the component of gravity along the incline: \[f = mg\sin\theta,\] directed up the incline.
Step 3: In time \(t\) the block moves with the elevator, so its displacement is purely vertical (upward) with magnitude \[d = vt.\]
Step 4: The friction force points up along the incline. The incline makes angle \(\theta\) with the horizontal, so the up-the-incline direction makes angle \(\theta\) with the horizontal, and its vertical component is \(\sin\theta\). The work done by friction is the dot product of the friction force and the vertical displacement: \[W_f = f \cdot d \cdot \sin\theta = (mg\sin\theta)(vt)(\sin\theta).\]
Step 5: Simplify: \[W_f = mgvt\sin^2\theta.\] \[\boxed{W_f = mgvt\sin^2\theta}\]