A 1000 N box lies on rough horizontal floor with coefficient of friction μ=3/k. The minimum possible applied force (its direction can be chosen) required to start motion is 600 N. Find k.
Physics · Mixed Chapter
JEE Main Practice

Question

A 1000 N box lies on rough horizontal floor with coefficient of friction μ=3/k. The minimum possible applied force (its direction can be chosen) required to start motion is 600 N. Find k.

G1 Physics · Accepted Answer

Step 1: A pulling force F at angle θ above horizontal gives N=W−Fsinθ at impending motion. Step 2: Limiting equilibrium Fcosθ=μ(W−Fsinθ)⇒ F=μW/(cosθ+μsinθ). Step 3: To minimize F, maximize the denominator: max(cosθ+μsinθ)=√(1+μ²). Thus Fmin=μW/√(1+μ²). Step 4: 600=1000μ/√(1+μ²)⇒0.36(1+μ²)=μ²⇒μ²=0.5625⇒μ=0.75. Step 5: 3/k=0.75⇒k=4.