Problem 53 - A Complete Resource Book in Physics for JEE Main

Question

A block of mass \(m\) is placed on the top of another block of mass \(M\) as shown in the Fig. \(3.81\). The co-efficient of friction between them is \(\mu .\) The maximum acceleration with which the block \(M\) may move so that \(m\) also moves along with it is (A) \(\mu g\) (B) \(g / \mu\) (C) \(\mu^{2} / g\) (D) \(g / \mu^{2}\)

Step-by-Step Solution

Answer

The correct short answer based on the provided step-by-step solution is: The maximum acceleration with which block \(M\) may move so that \(m\) also moves along with it is given by the relation: \[ a_{max} = \frac{\mu m g}{M} \]

Step 1: Identify the forces acting on the blocks

Two forces acting on each block. For block \(m\), there is a friction force \(f\) that opposes the motion, and its weight \(mg\) acting downwards. For block \(M\), there is an applied force \(F\) to the right, the friction force \(f\) acting in the opposite direction to the applied force and its weight \(Mg\) acting downward.

Step 2: Analyze the forces acting on block \(m\)

To ensure that block \(m\) moves along with block \(M\), the friction between the blocks must be enough to prevent block \(m\) from sliding. The maximum friction force that can act on block \(m\) is given by the formula: \[ f_{max} = \mu N = \mu m g \] Where \(N\) is the normal force acting on the block, which is equal to its weight \(mg\).

Step 3: Analyze the forces acting on block \(M\)

For block \(M\), we can write Newton's second law of motion. The sum of the forces in the x-direction is equal to the product of mass and acceleration. Let's denote the acceleration of block \(M\) as \(a\). Then we have: \[ F - f = M a \]

Step 4: Find the maximum acceleration of block \(M\)

Since we want the maximum acceleration, we substitute the maximum friction force \(f_{max}\) in the equation from step 3: \[ F - \mu m g = M a \] Now, we can solve for \(a\): \[ a = \frac{F - \mu m g}{M} \] However, in the problem, the applied force \(F\) isn't given, but since we want the maximum acceleration, the equality in the equation should hold: \( F = \mu m g \). Thus, we can find the maximum acceleration with which block \(M\) may move so that \(m\) also moves along with it by dividing the force by the mass \(M\): \[ a_{max} = \frac{\mu m g}{M} \] Now, let's analyze the answer choices to see which one matches our result. (A) \(\mu g\): This choice would be correct if \(M = m\), but we don't have any information about the relation between \(M\) and \(m\), hence not the correct option. (B) \(g / \mu\): Incorrect. The relationship between \(a_{max}\) and \(g\) is not an inverse one. (C) \(\mu^{2} / g\): Incorrect. There is no square in acceleration nor any inverse relationship between \(a_{max}\) and \(g\). (D) \(g / \mu^{2}\): Incorrect. Similar reasoning as choice (C). None of the given options match our result. The maximum acceleration should be given by the relation: \[ a_{max} = \frac{\mu m g}{M} \]