# Homework Help: Spring Forced Distance

1. Dec 24, 2009

### 312213

1. The problem statement, all variables and given/known data
A spring of length 0.80 m rests along a frictionless 30° incline (a). A 2.6 kg mass, at rest against the end of the spring, compresses the spring by 0.10 m. (b)
http://img64.imageshack.us/img64/5889/physics.jpg [Broken]

The mass is pushed down, compressing the spring an additional 0.60 m, and then released. If the incline is 2.0 m long, determine how far beyond the rightmost edge of the incline the mass lands.

The spring constant is 127.4

2. Relevant equations
Us=1/2kx²
v²=v0²+2ax
v=v0+at
x=x0+v0t+1/2at²

3. The attempt at a solution
0.8m, in a 30° angle, on the spring will count as the 0 point.
The potential energy of the spring on the block is 1/2kx²=Us so
(1/2)(127.4)(0.1+0.6)²=31.213J

Ki+Ui=Kf+Uf
0+31.213=Kf+0

K=1/2mv²
31.213=1/2(2.6)v²
4.9=v

http://img691.imageshack.us/img691/5889/physics.jpg [Broken]
Diagram of forces show gravity acts against it so mgsin30
(2.6)(-9.8)sin30=-12.74N
F=ma
-12.74=2.6a
-4.9=a

v²=v0²+2a(x-x0)
v²=4.9²+2(-4.9)(1.9)
2.32=v

So the block leaves the ramp at ~2.32m/s, 30° above the horizontal.

I use the time it takes for the block to reach its peak height and time to hit the ground and multiply that with the velocity it moves rightwards.

v=v0+at
0=2.32sin30+-9.8t
0.12=t1

In this time the block moves up so its final height is
x=x0+v0t+1/2at²
x=0+(2.32sin30)0.12+(1/2)(-9.8)(0.12)²
x=0.06875+1 (original height of right edge of ramp)
x=1.06875

Next is the time it takes to land,
x=x0+v0t+1/2at²
1.06875=0+0+(1/2)(9.8)t²
0.467=t2

t1+t2=0.585

Finally the distance it goes away from the edge is velocity times time so

x=vt
x=(2.32cos30)(0.585)
x=1.1771557879568892088679537911192

Is this right or are there something wrong with the way I approached the problem?

Last edited by a moderator: May 4, 2017
2. Dec 25, 2009

### Staff: Mentor

That's the elastic potential energy at the lowest point.

Don't forget gravitational potential energy.

Hint: Rather than treating the spring and gravity separately, use conservation of energy to solve for the speed of the mass as it leaves the incline.