# Determining the speed of a mass attached to a spring when stretched

1. Jun 10, 2012

### Vegeance

1. The problem statement, all variables and given/known data

3. A 0.5 kg mass is connected to the lower end of an ideal vertical coil spring suspended from a retort stand. The mass is held at rest with the spring completed unstretched and then allowed to fall. If the spring constant is 20 N/m, determine the maximum distance the mass drops before coming momentarily to rest and starting up again. Ignore frictional losses. (50 cm)

Egravity potential=Espring
[0.5][kx2]=mgx
[0.5][kx]=mg
[2mg/k]=[x]
[x] = 0.50m

4. In the problem above, the mass is now held initially at rest at a position where the spring is stretched 20 cm. Determine the maximum distance the mass will fall. (0.1 m)

Espring = Espring after - Egravity potential work done
[0.5][k(0.2)2] + [0.5kg][~10N/kg][x']=[0.5k][x+0.2m][2]
[0.4J]+[5x']=[10][x'2+0.4x'+0.04]
[0.4J]+[5x']=[10x'2-x'+0.4]
0=[10x2-x']
x = (1/10; 0)

5. In problem 3, determine how fast the mass was travelling on the way down when the spring was stretched 30 cm.
(1.5 m/s)

I am getting stuck at this question as how to conceptualize it, although I tried substituting to no avail.
Initial State : Potential Energy of Spring + Gravity Potential Energy Work Done
Final State : Potential Energy of Spring[x+0.3m] + Kinetic Energy

2. Relevant equations

Eg = mgh
Ek = 1/2mv^2
Es = 1/2kx^2

3. The attempt at a solution

2. Jun 10, 2012

### BruceW

It looks like you have done 3 and 4 correctly. For question 5, you can do it. You have written that the final state potential energy in the spring is at position (x+0.3m) What is x meant to be? The original position? If so, then they have told you in question 3 the original potential energy, so this gives you a value for x.