IIT JEE Advanced 2011 Paper 2

From conservation of energy, we have

1

2

mu2 = µmgx+

1

2

kx2
where x is the displacement of the block. Here, LHS is the energy supplied to the system, first term at RHS is the work done against friction and the last term is the energy stored in the spring.
x = 0.06 m, m = 0.18 kg, μ = 0.1 and k = 2 N/m. Therefore,
u2 = 2 × 0.1 × 10 × 0.06 + [2×

(0.06)2

0.18

] = 0.16 m/s
or u = 0.4 =

N

10

⇒ N = 4