A block of mass 30.0 kg is being brought down by a chain. If the block acquires a speed of 40.0 cm/s in dropping down 2.00 m, find the work done by the chain during the process.
The distance the block moves, s = 2 m
Mass of the block, m = 30 kg
Initially, the speed of the block, u = 40 cm/s = 0.4 m/s
Final speed of the block, v = 0 m/s
Work done on the block = change in kinetic energy of the block
= 1/2 × m × (v2-u2)
= 0.5 × 30 × [0 – (0.4×0.4)]
= -2.4 J
Let the tension produced in the string be T.
Net force on the block = Tension in the string - Weight of the block
F = T - mg
F = T - 30 × 9.8
F = T - 294
Also, work done on the block = Fs = (T-294) ×2
-2.4 = (T-294) × 2
T = 292.8 N
Work done on the block by this tension in the string = Tscos180°
(As angle between tension T and displacement s is 180°)
= 292.8 × 2 × -1
= -585.6 J
≈ -586 J