On top of a spiral spring of the force constant 500Nm−¹ is placed a mass of 5 x 10−³ kg. If the spring is compressed downwards by a length of 0.02m and then released; calculate the height to which the mass is projected?

A. 8 m
B. 4 m
C. 2 m
D. 1 m

Correct Answer:

Option C = 2 m

Explanation

K = 500Nm^-1; M = 5 x 10^-3kg; e = 0.02m;
g = 10ms^-2. The elastic potential energy stored in the compressed spring is given by = ¹/₂Ke², and the energy is transformed into gravitational potential energy = mgh, thus from the principle of energy conversion, we have that mgh = ¹/₂Ke²
5 x 10^-3 x 10 x h = ¹/₂ x 500 x (0.02)²
∴ h = 1 x 500 x 0.02 x 0.022 x 5 x 10^-3 x 10
= 5 x 2 x 0.02___10^-1
= 10 x 0.02 x 10 = 2
∴ h = 2m