mass = 0.500 kg
Force = m g = 0.5 * 9.81 = 4.905 Newtons
extension = 0.10 meters
so
k = 4.905 / 0.1 = 49.05 Newtons / meter
I do not see any further questions but anyway
additional PE stored in spring if pulled further 0.50 meter
Potential Energy in spring = (1/2) k x^2 = (1/2) 49.05 (.25)
= 6.13 Joules
that will be the KE as it passes through equilibrium static stretchpoint
(1/2) m v^2 = .5 * .5 * v^2 = 6.13
so max v = 4.95 m/s
A body of mass 500g suspended from the end of a spiral spring which obeys Hooke's law, produced an extension of 10cm. If the mass is pulled down a distance of 50cm and released, calculate
a) the force constant of the spring
1 answer
1 answer