A large slab of mass 5 kg lies on a smooth horizontal surface, with a block of mass 4 kg lying on the top of it; the coefficient of friction between the block and the slab is 0.25. If the block is pulled horizontally by a force of \(F = 6\) N, the work done by the force of friction on the slab between the instants \(t = 2\) s and \(t = 3\) s is
Show Hint
Always first check whether the blocks move together or separately by comparing applied force with maximum static friction. Then use work-energy methods.
Step 1: Understanding the Concept: Check if both move together. Maximum friction \(= \mu mg = 0.25\times4\times10 = 10\) N \(> F=6\) N, so they move as one.
Step 2: Detailed Explanation: Combined acceleration: \(a = 6/9 = 2/3\) m/s\(^2\). Friction on slab: \(f = 5 \times a = 10/3\) N. Displacements: \(S(2) = \frac{1}{2}\times\frac{2}{3}\times4 = 4/3\) m; \(S(3) = \frac{1}{2}\times\frac{2}{3}\times9 = 3\) m. Displacement in interval: \(\Delta S = 3 - 4/3 = 5/3\) m. Work done by friction on slab: \[ W = f\,\Delta S = \frac{10}{3}\times\frac{5}{3} = \frac{50}{9} = 5.55 \text{ J} \]
Step 3: Final Answer: Work done by friction on slab \(= 5.55\) J.
