Physics help please?

Question

A 3.00-kg mass is fastened to a light spring that passes over a pulley. They pulley is frictionless, and its inertia may be neglected. The mass is released from rest when the spring is unstretched. If the mass drops 10.0cm before stopping, find (a) the spring constant of the spring and (b) the speed of the mass when it is 5.00cm below its starting point.

i got part A..but i dont understand how to do B..so can someone please teach/show me how to do it. that would be great.

Answer 1

At 5.00 cm, the spring is extended 1/2 of it's fully extended length. The energy at this state is k(.5x)^2 = 1/4 kx^2, that is, one-fourth of the energy in the spring when fully extended - a word of caution.
I used the kinematic approach for fun.
Sum F = T - mg = kx -mg = ma
So, a = (k/m)x - g, a function of x.

Now a = dv/dt = (dv/dx)(dx/dt) = v dv/dx.
Then a dx = v dv, and Int[ a dx ] = Int[ v dv ]
Substituting for a:
Int[ (k/m x - g) dx] = Int[ v dv ]
or k/(2m)x^2 gx ] from 1 to 2 = v2 ^2 - v1^2

Using k= 294 kg/s^2, I got v2 = .606 m/s at 5.00 cm.
I had to change the direction of the + coordinate to be positive downward for the mass to avoid taking the square root of a negative quantity to get v2.

Answer 2

I don't have a clear picture of this setup. But if you calculated the spring constant you can calculate the energy in the spring when it is stretched 10 cm.

Energy = F*d

At 5 cm extension the spring energy is 1/2 what it is at 10 cm. The other 1/2 of that energy is kinetic. That should be all the clue you need.