Asked by Fynface
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
a) the force constant of the spring
Answers
Answered by
Anonymous
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.