Two masses m₁ and m₂ are connected by a light spring as shown in the figure. Blocks are placed on a rough horizontal surface. Coefficients of friction μ₁ and μ₂ are given for each surface. A constant horizontal force F₀ is applied on m₂ as shown in the figure. If the minimum value of F₀ so that m₁ will just move is K, then the value of K/10 will be? Take m₁ = m₂ = 10 kg; μ₁ = μ₂ = 0.2.
Physics · Mixed Chapter
JEE Mains

Question

Two masses m₁ and m₂ are connected by a light spring as shown in the figure. Blocks are placed on a rough horizontal surface. Coefficients of friction μ₁ and μ₂ are given for each surface. A constant horizontal force F₀ is applied on m₂ as shown in the figure. If the minimum value of F₀ so that m₁ will just move is K, then the value of K/10 will be? Take m₁ = m₂ = 10 kg; μ₁ = μ₂ = 0.2.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Maximum static friction on each block is μmg=0.2×10×10=20 N. Block m₁ remains at rest until the spring force first reaches 20 N. Step 2: While m₂ slides but m₁ remains at rest, the effective constant drive on m₂ is F₀−20 N, opposed by the spring. Starting from zero extension, its maximum compression follows from work-energy: (F₀−20)xₘₐₓ=(1/2)kxₘₐₓ². Step 3: Maximum spring force is kxₘₐₓ=2(F₀−20). At the threshold for m₁ motion set 2(F₀−20)=20. Step 4: F₀=30 N=K, so K/10=3.