- #1

NewSoul

- 14

- 0

## Homework Statement

Here is an illustration to ease understanding:

Angle inclined plane: θ = 20.0°

spring of force constant k = 525 N/m

block: m = 2.71 kg

distance of block from spring: d = 0.330 m

speed of block: v = 0.750 m/s

By what distance is the spring compressed when the block momentarily comes to rest? (I'm going to call it 'x' in this problem.)

The answer should be x = 0.143 m

## Homework Equations

Ok, I'm not quite sure how many of these are relevant or if I'm missing some, but I'll just give some that I know.

PE

_{s}= (1/2)kx

^{2}(x = distance spring is compressed)

PE

_{g}= mgh

KE = (1/2)mv

^{2}

F = ma

U + KE = 0 (U = any potential energy? All potential energies? I'm not quite positive what is going on here.)

## The Attempt at a Solution

Okay. I'm not quite sure how to go about this, so I'll just start doing what I seem to know.

Let's start with force. It doesn't seem inherently relevant to the problem, but why not?

Okay, so force in the x direction seems to be: mgsinθ = ma. So we get a = gsinθ = (9.8)(sin 20°) = 3.35198 m/s

^{2}.

Force in the y direction seems to be: N - mgsinθ = 0. So N = mgsinθ = (2.71)(9.8)(sin 20°) = 9.08337 N.

-------------

Ok. That was probably not relevant. Let's do work now.

Using the numbers above, I get KE = (1/2)(2.71)(.750)

^{2}= 0.76219 J.

PEg = (2.71)(9.8)h. Okay, I'm not quite sure if h should just be d or if I have to factor in the slope. I'm thinking that slope probably does have to be included, so h is 0.330(sin 20°), so PE

_{g}is 2.99751 J.

So it seems like PE

_{g}+ PE

_{s}+ KE = 0. Then it must be that (1/2)kx

^{2}+ mgh + (1/2)mv

^{2}= 0... However, solving for x gives me the square root of a negative number, I must be interpreting U + KE = 0 wrong. Where do I go from here? Thanks, I know this is long.