Given m = 20 × 10-3kg
u = 60ms-1 & v = 30 ms-1
work done = decrease in kinetic energy
= \(\frac {1}{2}\)m (u2 – v2)
= \(\frac {1}{2}\)× 0.02 (602 – 302)
= 27J
From, v = u + at
30 = 60 + a (0.01)
a =\(\frac {-30}{0.01}\) = – 3000 ms-2
Retarding force, F = ma
= 0.02 × 3000 = 60N
work done, W = F × S
27 = 60 × S
∴ Thickness of the plank, S = \(\frac {27}{60}\)
= 0.45 m.