Question

a block of mass m slides on the wooden wedge which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is: 
[ Given m= 8kg, M=16kg ]
[Assume all the surfaces shown in the figure to be frictionless] 
 

• A. \(\frac{4}{3}g\)
• B. \(\frac{6}{5}g\)
• C. \(\frac{3}{5}g\)
• D. \(\frac{2}{3}g\)

Answer 1

The Correct Option is D

To solve the problem, we need to find the acceleration of the block with respect to the wedge. The given problem states that all surfaces are frictionless and there are two masses involved: the block of mass \( m = 8 \, \text{kg} \) and the wedge of mass \( M = 16 \, \text{kg} \).

Let's analyze the forces acting on the block and the wedge:

1. The block of mass \( m \) will experience a gravitational force downward, which is \( mg \), where \( g \) is the acceleration due to gravity.
2. Due to the incline of the wedge, this gravitational force can be resolved into two components:
   • Parallel to the incline: \( mg \sin \theta \)
   • Perpendicular to the incline: \( mg \cos \theta \)
3. Since all surfaces are frictionless, the normal force exerted by the wedge on the block is \( N = mg \cos \theta \).

We use Newton’s laws to find the accelerations:

For the block on the incline: F_{\parallel} = m a_{\mathrm{block}} = mg \sin \theta - N \sin \theta.

For the wedge on the horizontal surface: F_{\mathrm{wedge}} = M a_{\mathrm{wedge}} = N \sin \theta.

The perpendicular components cancel out, so: mg \sin \theta = m a_{\mathrm{block}} + M a_{\mathrm{wedge}}\cdot \cos \theta.

Simplifying with \(\sin 30^\circ = \frac{1}{2}\) and \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), we get: mg \cdot \frac{1}{2} = m a_{\mathrm{block}} + M a_{\mathrm{wedge}} \cdot \frac{\sqrt{3}}{2}.

Solving, we get: a_{\mathrm{block}} = \frac{2}{3}g.

This matches the correct option:

\(\frac{2}{3}g\)