Question

on top of a spiral spring of force constant 509N is placed placed in mass of 5x10raise to power 3 . If the spring is Compressed down ward by alength of 0.02m and then release Calculate the height to which the mass projected 

Answer 1

According to the energy conservation, the initial potential energy of the spring is equal to the final gravitational potential energy of the body:

$$\dfrac{kx^2}{2} = mgh$$

where $k = 509N/m, x = 0.02m, g = 9.8m/s^2, m=5000kg$ and $h$ is required height. Thus, obtain:

$$h = \dfrac{kx^2}{2mg} = \dfrac{509\cdot 0.02^2}{2\cdot 5000\cdot 9.8} \approx 2\times 10^{-9}m$$