Asked by Jess
a 1 kg mass is attached to a spring hanging vertically and hangs at rest in the equilibrium position. the spring constant of the spring is 1 n/cm. the mass is pulled downward 2 cm and released. what is the speed of the mass when it is 1 cm above the point from which it was released?
Answers
Answered by
drwls
Max spring potential energy
= (1/2) k X^2 = (100/2)*(0.02)^2
= 0.2 Joules
When 0.01 m below the release point, the spring has 1/4 of its max potential energy, so 0.15 J is available for kinetic energy
(1/2)MV^2 = 0.15 J can be solved for the speed V. I get 0.548 m/s
It is permissible to neglect gravity in this problem when you regard kx as the force applied by the spring, including Mg, with x measured from the equilibrium position. g does not affect the answer or the period of vibration.
= (1/2) k X^2 = (100/2)*(0.02)^2
= 0.2 Joules
When 0.01 m below the release point, the spring has 1/4 of its max potential energy, so 0.15 J is available for kinetic energy
(1/2)MV^2 = 0.15 J can be solved for the speed V. I get 0.548 m/s
It is permissible to neglect gravity in this problem when you regard kx as the force applied by the spring, including Mg, with x measured from the equilibrium position. g does not affect the answer or the period of vibration.
Answered by
arwa
the answer is 1.73 but i don't know how can every one help me ??
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.