Question

A block P of mass m is placed on a frictionless horizontal surface. Another block Q of same mass is kept on P and connected to the wall with the help of a spring of spring constant k as shown in the figure. $\mu_s$ is the coefficient of friction between P and Q. The blocks move together performing SHM of amplitude A. The maximum value of the friction force between P and Q is

• A. kA
• B. $\frac{kA}{2}$
• C. zero
• D. $\mu_s$mg

Answer 1

Answer: (B)

$\frac{kA}{2}$

Answer 2

To determine the maximum value of the friction force between blocks P and Q, we first need to find the maximum acceleration of the system.

1. Consider the combined system of blocks P and Q:

   Since the blocks move together, we can treat them as a single system with total mass M_total = m_P + m_Q = m + m = 2m. The only horizontal external force acting on this combined system is the spring force. The spring force is given by F_spring = -kx, where k is the spring constant and x is the displacement from the equilibrium position. For Simple Harmonic Motion (SHM) with amplitude A, the maximum displacement is A. Therefore, the maximum spring force is F_max_spring = kA.

2. Calculate the maximum acceleration of the system:

   According to Newton's second law, F_net = M_total * a. The maximum acceleration a_max occurs when the spring force is maximum: F_max_spring = M_total * a_max kA = (2m) * a_max Therefore, the maximum acceleration of the combined system is: a_max = \frac{kA}{2m}

3. Analyze the forces on block P:

   Block P is placed on a frictionless horizontal surface. The only horizontal force acting on block P is the static friction force exerted by block Q on P. This friction force is what causes block P to accelerate along with block Q. Let f be the friction force. According to Newton's second law for block P: f = m_P * a f = m * a

4. Determine the maximum friction force:

   The friction force f will be maximum when the acceleration a is maximum. Substitute the value of a_max into the equation for f: f_max = m * a_max f_max = m * \left(\frac{kA}{2m}\right) f_max = \frac{kA}{2}

This is the maximum value of the friction force between P and Q. The coefficient of static friction μ_s is given, which implies that for the blocks to move together without slipping, the required maximum friction force (kA/2) must be less than or equal to the maximum possible static friction (μ_s mg). However, the question asks for the value of the friction force itself, assuming they do move together.

The final answer is $\frac{kA}{2}$.

Explanation of the solution:

The blocks P and Q move together, so their combined mass is $2m$. The maximum force exerted by the spring is $kA$. This force causes the combined system to accelerate. The maximum acceleration of the system is $a_{max} = \frac{F_{max}}{M_{total}} = \frac{kA}{2m}$. Block P, of mass $m$, is accelerated solely by the friction force from block Q. Therefore, the maximum friction force $f_{max}$ required to accelerate block P is $f_{max} = m \cdot a_{max} = m \cdot \frac{kA}{2m} = \frac{kA}{2}$.