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: C
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
