Problem 100 - A Complete Resource Book in Physics for JEE Main

Question

A block \(P\) of mass \(4 \mathrm{~kg}\) is placed on horizontal rough surface with co-efficient of friction \(\mu=0.6\). And two blocks \(R\) and \(Q\) of masses \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) connected with the help of massless strings \(A\) and \(B\), respectively, passing over frictionless pulleys as shown, then \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) Acceleration of block \(P\) is zero. (B) Tension in string \(A\) is \(20 \mathrm{~N}\). (C) Tension in string \(B\) is \(40 \mathrm{~N}\). (D) Contact force on block \(P\) is \(20 \sqrt{5} \mathrm{~N}\).

Step-by-Step Solution

Answer

The short answer is: The acceleration of block P is \(9\,\frac{\text{m}}{\text{s}^2}\), tension in string A is \(20\,\text{N}\), and tension in string B is \(40\,\text{N}\). Therefore, statements B and C are true, while statements A and D are false.

Step 1: Analyze the forces acting on each block

First of all, we are going to analyze the forces acting on each block. For block P, there is a gravitational force W = mg acting downward and a normal force N acting upward, and a friction force f = μN acting horizontally to the left. For blocks Q and R, there is a gravitational force acting downward and the tension in the strings (TA and TB) acting upward.

Step 2: Calculate the maximum friction force for block P

Since block P is on a horizontal surface, the normal force N is equal to the gravitational force W for this block: \(N = W = mg = (4\,\text{kg})(10\,\frac{\text{m}}{\text{s}^2}) = 40\,\text{N}\) Now we'll calculate the maximum friction force that can act on block P: \(f_\text{max} = \mu N = (0.6)(40\,\text{N}) = 24\,\text{N}\)

Step 3: Calculate the net force acting on block Q

There are two forces acting on block Q: the gravitational force and the tension in string A. So we can write the following equation: \(T_A = m_Qg \Rightarrow T_A = (2\,\text{kg})(10\,\frac{\text{m}}{\text{s}^2}) = 20\,\text{N}\)

Step 4: Calculate the net force acting on block R

There are also two forces acting on block R: the gravitational force and the tension in string B. So we can write the following equation: \(T_B = m_Rg \Rightarrow T_B = (4\,\text{kg})(10\,\frac{\text{m}}{\text{s}^2}) = 40\,\text{N}\)

Step 5: Calculate the net force acting on block P and its subsequent acceleration

For block P, there is a horizontal net force acting to the right with a magnitude equal to the tensions in the strings: \(F_\text{net} = T_A + T_B = 20\,\text{N} + 40\,\text{N} = 60\,\text{N}\) Since the net force on block P (60 N) is greater than the maximum friction force (24 N), the block will accelerate and not remain stationary. Let's now find the acceleration or block P: \(a_P = \frac{F_\text{net} - F_\text{friction}}{m} = \frac{60\,\text{N} - 24\,\text{N}}{4\,\text{kg}} = \frac{36\,\text{N}}{4\,\text{kg}} = 9\,\frac{\text{m}}{\text{s}^2}\)

Step 6: Check the given statements

Now, let's verify if the given statements are true or false based on our calculations: (A) Acceleration of block P is zero. [False] - We found that the acceleration of block P is 9 m/s². (B) Tension in string A is 20 N. [True] - We calculated the tension in string A to be 20 N. (C) Tension in string B is 40 N. [True] - We calculated the tension in string B to be 40 N. (D) Contact force on block P is \(20\sqrt{5}\,\text{N}\). [False] - The contact force is the normal force N, which is equal to 40 N, not \(20\sqrt{5}\,\text{N}\). In conclusion, statements B and C are true, while statements A and D are false.