A block of mass 3

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