How Does Friction Affect the Velocity of a Mass on a Spring?

[henry2221]

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?

 

[shawshank]

all the elastic potential energy is converted back into kinetic. so equate Eelastic - Ethermal = Ekinetic and solve. but you have to also consider energy lost due to friction, so put that in the equation too.

 

[henry2221]

all the elastic potential energy is converted back into kinetic. so equate Eelastic - Ethermal = Ekinetic and solve. but you have to also consider energy lost due to friction, so put that in the equation too.

wait... is this assuming that I did part a correctly? or...

 

[Astronuc]

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.

Part a must include the work associated with the frictional force of 5 N over 1 m, in addition to the mechanical energy in the spring.

In part b, when the spring recoils, the friction is again present, so not all the spring energy will be transformed into kinetic energy.

 

[henry2221]

So I am assuming that its:

W = Us + Ug + 5Nm

However how would I find Ug without knowing it's height?

 

[Astronuc]

The mass changes elevation by 1 m, from the equilibrium position, and since the mass is going down, it's gravitational potential energy (GPE) decreases. When the mass goes up, it's GPE increases.

 

[henry2221]

New attempt:
Us = W = U = (1/2)k x^2
(1/2)(10 N/m) (1m)^2
= 5 N*m

Ug = (0.1 kg)(1m)(-9.8 m/s^2)
= -.98 J


W= Us + Ug+ (-5J)
5J + (-98 J) + (-5J)
= -.98 J

b) W = E - Wnc (Work energy Theorem for Systems)
W = 0 Because there is no internal nonconservative forces.. I still haven't a clue where to start with this one...
Therefore E:

 

[Astronuc]

OK, when the spring is being pulled down, the friction and spring force act in the same direction. Part of the work being done goes into the stored mechanical energy in the spring and part is used to overcome friction.

When the spring recoils, the spring pulls (acts up) but friction is acting down.

 

[henry2221]

New attempt:

Us = W = U = (1/2)k x^2
(1/2)(10 N/m) (1m)^2
= 5 N*m

Ug = (0.1 kg)(1m)(-9.8 m/s^2)
= -.98 J

Uf = 5J


W= Us - Ug + (5J)
5J -.98 J + 5J
= 9.8 J

So far, is this correct?