(adsbygoogle = window.adsbygoogle || []).push({}); 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=U_{s}so

(1/2)(127.4)(0.1+0.6)²=31.213J

K_{i}+U_{i}=K_{f}+U_{f}

0+31.213=K_{f}+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=v_{0}²+2a(x-x_{0})

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=v_{0}+at

0=2.32sin30+-9.8t

0.12=t1

In this time the block moves up so its final height is

x=x_{0}+v_{0}t+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=x_{0}+v_{0}t+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?

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# Homework Help: Spring Forced Distance

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