- #1
henry2221
- 20
- 0
Question:
A 0.2 kg mass is attached to a spring k = 10 N/m) and hangs vertically near the Earth's surface (g = 9.81 m/s2 ). The mass makes contact with a wall as it moves vertically and a constant frictional force of magnitude 5N acts on the mass as it moves.
a) Calculate the amount of work required to pull the spring down by 1 m.
b) Calculate the speed of the mass as it passes through the equilibrium position after being pulled down by 1 m.
Eq'n
U = (1/2)k x^2
Attempt:
a) W = U = (1/2)k x^2
(1/2)(10 N/m) (1m)^2
= 5 N*m
= 5 J
I am lost on part b, someone suggested
v = sqrt(k/m) * x
... But I have no clue where they derived this equation...
I tried
K = 1/2 m v^2
v = sqrt(2K/m)
...But I believe this is incorrect... suggestions?
A 0.2 kg mass is attached to a spring k = 10 N/m) and hangs vertically near the Earth's surface (g = 9.81 m/s2 ). The mass makes contact with a wall as it moves vertically and a constant frictional force of magnitude 5N acts on the mass as it moves.
a) Calculate the amount of work required to pull the spring down by 1 m.
b) Calculate the speed of the mass as it passes through the equilibrium position after being pulled down by 1 m.
Eq'n
U = (1/2)k x^2
Attempt:
a) W = U = (1/2)k x^2
(1/2)(10 N/m) (1m)^2
= 5 N*m
= 5 J
I am lost on part b, someone suggested
v = sqrt(k/m) * x
... But I have no clue where they derived this equation...
I tried
K = 1/2 m v^2
v = sqrt(2K/m)
...But I believe this is incorrect... suggestions?