Problem 41 - A Complete Resource Book in Physics for JEE Main

Question

With what minimum acceleration mass \(M\) must be moved on frictionless surface so that \(m\) remains stick to it as shown. The co-efficient of friction between \(M\) and \(m\) is \(\mu\). (A) \(\mu g\) (B) \(\frac{g}{\mu}\) (C) \(\frac{\mu m g}{M+m}\) (D) \(\frac{\mu m g}{M}\)

Step-by-Step Solution

Answer

(D) \(\frac{\mu mg}{M}\)

Step 1: Identify the forces acting on the masses

: For mass \(M\), there is a normal force \(N\) acting perpendicular to the surface, the weight \(Mg\) acting downward, and the static friction force \(F_s\) acting on the opposite direction of the acceleration. For mass \(m\), there is a normal force \(N'\) acting perpendicular to the surface of \(M\), the weight \(mg\) acting downward, and the static friction force \(F_s\) acting horizontally.

Step 2: Write down Newton's second law equation for each mass

: Newton's second law states that the net force acting on an object is equal to the product of its mass and acceleration: \(F_{net} = ma\). For mass \(M\), the net force in the horizontal direction is \(F_s = M a\). For mass \(m\), the net force in the vertical direction is \(N' - mg = 0\) and in the horizontal direction is \(F_s = m a'\).

Step 3: Establish the relationship between accelerations of masses \(M\) and \(m\) and the static friction force \(F_s\)

: Considering the friction and the accelerations, we can write: \(F_s = M a = m a'\) Since we are looking for the minimum acceleration, the static friction force \(F_s\) will be at its maximum value, which is given by: \(F_s = \mu N'\) We know that for mass \(m\), in the vertical direction, the net force is zero. Therefore, \(N' = mg\) Now, we can substitute this to find the static friction force: \(F_s = \mu mg\)

Step 4: Solve for the minimum acceleration of mass M

: Now we have the static friction force \(F_s\) at its maximum value, we can substitute this in the equation (1) for mass \(M\): \(F_s = M a\) \(\mu mg = M a\) We can now solve for \(a\): \(a = \frac{\mu mg}{M}\) Therefore, the minimum acceleration of mass \(M\) required is

Step 5: Answer

: (D) \(\frac{\mu mg}{M}\)