On top of a spiral spring of the force constant... - JAMB Physics 2003 Question
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 option: c
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
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
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