The blocks of masses m₁ = 1 kg and m₂ = 2 kg are connected by a spring and rest on a horizontal surface. The spring is unstretched initially. The spring constant of the spring is k = 8 N/m. The coefficient of friction between blocks and horizontal surface is μ = 1/2. Now the left block is imparted a velocity u towards right as shown in figure. Then what is the largest value of u such that the block mass m₂ never moves?
Physics · Mixed Chapter
JEE Mains

Question

The blocks of masses m₁ = 1 kg and m₂ = 2 kg are connected by a spring and rest on a horizontal surface. The spring is unstretched initially. The spring constant of the spring is k = 8 N/m. The coefficient of friction between blocks and horizontal surface is μ = 1/2. Now the left block is imparted a velocity u towards right as shown in figure. Then what is the largest value of u such that the block mass m₂ never moves?

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: For m₂ to remain at rest, the spring force must not exceed its limiting static friction, Fₛ = μm₂g = (1/2)(2)(10) = 10 N. Step 2: The greatest compression is xₘₐₓ = Fₛ/k = 10/8 = 1.25 m. Step 3: During compression of m₁, kinetic energy is stored in the spring and lost against friction: (1/2)m₁u² = (1/2)kxₘₐₓ² + μm₁gxₘₐₓ. Step 4: (1/2)u² = 6.25 + 6.25 = 12.5, hence u = 5 m/s. Answer: 5.