# Spring equilibrium problem

• starhallie

## Homework Statement

A 2kg object is suspended from a vertical spring that has a constant of 180N/m.

a) How far does the spring stretch from the unstrained length?

b) If the object is now pulled downward an additional distance of 5cm. What external force was required to do so?

c) When the object is released from rest, it begins to move upwards. Find the maximum speed that the object reaches and the maximum height above the release point at which the object releases direction.

Fspring= kx

## The Attempt at a Solution

I would be grateful if someone could check my answers to parts a & b and possibly steer me in the right direction on part c?

a) Fspring= kx
F=ma
ma=kx
x= ma/k
=(2kg)(9.8 m/s2)/(180 N/m)
= 0.1089m

b) Fspring= kx
=(180 N/m)(0.05cm)
= 9 N

In the case b, find the elastic potential energy stored in the spring.
After releasing when it crosses the position in case a, it will have the maximum kinetic energy and gravitational potential energy. Apply the conservation of energy to find the maximum velocity.

Thanks for your help This is what I got so far...

PEspring= 1/2kw2= 1/2(180 N/m)(0.05m)= 0.225 J

... but I'm having trouble understanding the rest.

At a position total energy = 1/2*m*v^2 + mgh
Find vmax

Thank you very much for your help!