# Horizontal Spring-Mass Problem

1. Oct 29, 2010

### loganblacke

1. The problem statement, all variables and given/known data
A 0.766-kg mass is attached to a horizontal spring with k = 78.0 N/m. The Mass slides across a frictionless surface. The spring is stretched 25.53 cm from equilibrium, and then the mass is released from rest.
(a) Find the mechanical energy of the system.
(b) Find the speed of the mass when it has moved 3.63 cm.
(c) Find the maximum speed of the mass.

2. Relevant equations
Fs=-kx
K=(1/2)mv^2
Work = Integral( Kx) dx from a to b

3. The attempt at a solution
F=-(78.0 N/m * .2553 m) = 19.9134 N
W=(78x^2)/2 from 0 to .2553 = 2.54 J

I'm not sure where to go from here..

2. Oct 29, 2010

### loganblacke

After I posted this a light went on and I used F=ma to find a = 25.99 m/s. I then used Vf^2 = Vi^2 + 2ad to find the answer to b and c, however I still got part c wrong for some reason. for part b I got Vf = sqrt(2(25.99)(.0363))=1.37. For part c I got Vf = sqrt(2(25.99)(.2553))=3.64 m/s/s. According to the webwork the correct answer to part c is 2.58. Does anyone know what I am doing wrong? Thanks!

3. Oct 29, 2010

### MichalXC

Work done by a spring is W=-(1/2)*k*x^2 from x_initial to x_final. Work done is also equal to the change in kinetic energy (by the work-energy theorem). These two pieces of information should enable you to find the velocity of the mass after it has moved 3.63 cm.

I hope this helps.