Problem 152 - A Complete Resource Book in Physics for JEE Main

Question

A block of mass \(M\) is pulled along a horizontal frictionless surface by a rope of mass \(m\). If a force \(P\) is applied at the free end of the rope, the force exerted by the rope on the block is (A) \(\frac{P m}{M+m}\) (B) \(\frac{P m}{M-m}\) (C) \(P\) (D) \(\frac{P M}{M+m}\)

Step-by-Step Solution

Answer

\(F = \frac{P M}{M+m}\)

Step 1: Analyze forces on the rope and block separately

As we are dealing with two separate objects (the rope and the block), we need to analyze the forces acting on each of them separately. We will first consider the rope and then the block.

Step 2: Apply Newton's second law for the rope

For the rope, the applied force P acts on it, and a force F (the force exerted by the rope on the block) acts in the opposite direction. According to Newton's second law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration. So, for the rope we have: \(P - F = m a\)

Step 3: Apply Newton's second law for the block

For the block, the only horizontal force acting on it is the force exerted by the rope (F). Therefore, Newton's second law for the block is: \(F = M a\)

Step 4: Solve for the force exerted by the rope on the block (F)

Now, we have two equations with the force F and the acceleration a: 1. \(P - F = m a\) 2. \(F = M a\) We can solve for F by eliminating the variable a. To do this, first, solve equation 2 for a, getting: \(a = \frac{F}{M}\) Now substitute this expression for a into equation 1: \(P - F = m (\frac{F}{M})\) Now solve for F. First, multiply both sides by M, yielding: \(P M - F M = m F\) Next, move the term with F to one side of the equation: \(P M = F(M + m)\) Finally, divide both sides by (M + m) to get F: \(F = \frac{P M}{M+m}\) So, the force exerted by the rope on the block is: \(F = \frac{P M}{M+m}\) Therefore, the correct answer is: (D) \(\frac{P M}{M+m}\)