Question

A 4 kg roller is attached to a massless spring of spring constant k = 100 n/m. It rolls without slipping along a frictionless horizontal road. The roller is displaced from its equilibrium position by 10 cm and then released. Its maximum speed will be (2012) (a) 0.5 m s-1 (b) 0.6 m s-1 (c) 0.4 m s-1 (d) 0.8 m s-

Answer 1

From the law of energy conservation we have,

Potential energy within string will be converted into kinetic energy.

(KX^2)/2 = (mv^2)/2

K = spring constant (Newton/m)

X = displacement (m)

m = Mass of roller (Kg)

V = Max. speed in m/s

From the question, we know

K = 100 Newton/m

X = 10cm = 1/10 m

m = 4 kg

V = ?

Substituting we get,  

100/(10*10) = 4*V^2

V = √(1/4) = ½ = 0.5m/s

Maximum speed achived is 0.5 m/s