Problem 149 - A Complete Resource Book in Physics for JEE Main

Question

One end of massless rope, which passes over a massless and frictionless pulley \(P\) is tied to a hook \(C\) while the other end is free. Maximum tension that the rope can bear is \(840 \mathrm{~N}\). With what value of maximum safe acceleration (in \(\mathrm{ms}^{-2}\) ) can a man of \(60 \mathrm{~kg}\) climb on the rope? (A) 16 (B) 6 (C) 4 (D) 8

Step-by-Step Solution

Answer

The maximum safe acceleration with which the man can climb the rope without exceeding the maximum tension is approximately 4 \(ms^{-2}\).

Step 1: Analyze the forces acting on the man while climbing

When the man is climbing the rope, his weight will be acting downwards, and the tension in the rope will be acting upwards. We can represent this by drawing a free body diagram for the man. We can also represent the force of tension acting on the man, due to the massless rope. We will assume that the man is climbing up the rope with an acceleration "a."

Step 2: Setting up the equation

Using Newton's second law of motion, we can write an equation for the forces acting on the man. \(T - mg = ma\) Where: T is the tension in the rope (in Newtons) m is the mass of the man (60 kg) g is the acceleration due to gravity (\(9.8 ms^{-2}\)) a is the acceleration of the man while climbing (in \(ms^{-2}\))

Step 3: Solve the equation for acceleration "a"

As we need to find the maximum safe acceleration, we need to ensure that the tension in the rope should not exceed the maximum tension it can bear (840 N). Maximum tension in the rope (T) should be equal to 840 N. Therefore, \(840 - 60 * 9.8 = 60 * a\)

Step 4: Calculate the maximum safe acceleration "a"

Now, we will simply solve the equation for "a": \(a = \frac{840 - (60 * 9.8)}{60}\) \(a = \frac{840 - 588}{60}\) \(a = \frac{252}{60}\) \(a = 4.2 ms^{-2}\) Since 4.2 is very close to 4, we can conclude that the maximum safe acceleration with which the man can climb the rope without exceeding the maximum tension is approximately 4 \(ms^{-2}\). Therefore, the correct answer is option (C).