A block of mass 25 kg is pulled along a horizontal rough surface by a force F making 45° with the horizontal. The coefficient of friction is 0.25. The block moves with uniform velocity. The work done by the applied force over a horizontal displacement of 5 m is:
Options

Question

A block of mass 25 kg is pulled along a horizontal rough surface by a force F making 45° with the horizontal. The coefficient of friction is 0.25. The block moves with uniform velocity. The work done by the applied force over a horizontal displacement of 5 m is:

Options

G1 Physics · Accepted Answer

Since the block moves with uniform velocity, the horizontal force balance is Fcos45° = f. Normal reaction is reduced by the upward component of F: N = mg - Fsin45°. Limiting/kinetic friction is f = μN = μ(mg - Fsin45°). Since sin45° = cos45°, let X = F/√2. Then X = μ(mg - X). Using μ=0.25 and g=9.8 m/s²: mg = 25×9.8 = 245 N. Thus, X = 0.25(245-X) ⇒ 1.25X = 61.25 ⇒ X = 49 N. Work done by the applied force is W = Fd cos45° = Xd = 49×5 = 245 J. Correct option: B.