How to calculate work done on spring from object on slope?

[Eclair_de_XII]

Homework Statement

"In Fig. 8-40, a block of mass $m=12kg$ is released from rest on a friction-less incline of angle $\theta=\frac{\pi}{6}$. Below the block is a spring that can be compressed $\frac{1}{50}m$ by a force of $270 N$. The block momentarily stops when it compresses the spring by $\frac{11}{200}m$. (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring?"

Homework Equations

$U_{spring}=\frac{1}{2}kx^2$
$W=\frac{1}{2}m(v^2-v_0^2)$

The Attempt at a Solution

$m=12kg$
$\theta=\frac{\pi}{6}$
$x=\frac{1}{50}m$
$F=270N$
$s=\frac{11}{200}m$

(a)
To find the distance that the block slides down, $d+x$, I started out by finding the spring constant $k$.

$F=-kx$
$k=-\frac{F}{x}=-(50m^{-1})(270N)=13,500\frac{N}{m}$

And then I state the potential energy in the spring and equate it to the gravitational work.

$U_{spring}=\frac{1}{2}ks^2=K_g=F⋅d=[(mg)][(cos(\frac{\pi}{2}-\theta))(d)]$
$d=\frac{sec(\frac{\pi}{2}-\theta)}{2mg}(ks^2)=\frac{10s^2}{(12kg)(98m)}(13,500\frac{N}{m})(\frac{121}{40000}m^2)=(\frac{1s^2}{294m⋅kg})(\frac{27}{16})(121J)=\frac{3267}{4704}m$

Then add x...

$d+x=\frac{606,656}{940,800}m$

I assumed it wanted vertical distance from its position at rest, so the distance I'm looking for is:

$(sin\theta)(d+x)=\frac{606,656}{1,881,600}m≈0.32m$

I feel like I'm misunderstanding this question, though; when I remove the x from the equation, I get the right answer... Please correct me if I'm wrong in assuming that they are looking for the distance the block has covered with or without the distance covered when compressing the spring. I don't feel like I will be ready to move onto part (b) until I have made sure that I have done everything in (a) correctly.

But if anyone was curious, I ended up using the distance part (a) was looking for, and using the SUVAT equation to solve for $v_f$. I also used it by equating $W_g=\frac{1}{2}mv^2$. Either way, I ended up with $v=\sqrt{2g(cos(\frac{\pi}{2}-\theta))(d)}$, and I don't think it's correct.

 

[haruspex]

How far has the block moved in losing its PE when it comes momentarily to rest?

 

[Eclair_de_XII]

It's moved $(d+x)(sin\theta)$ in terms of vertical distance, I'm guessing?

 

[haruspex]

It's moved $(d+x)(sin\theta)$ in terms of vertical distance, I'm guessing?

Right. But you used $mg(d sin\theta)$ for lost PE in post #1, yes?

 

[Eclair_de_XII]

But you used mg(dsinθ)mg(d sin\theta) for lost PE in post #1, yes?

Right. I assumed that gravity stopped doing work when the block touched the spring. So I'm thinking it's actually:

$U_{spring}=(mg)(sin\theta)(d+x)$
$(d+x)=\frac{3267}{4704}m$

And the distance I'm looking for is:

$(sin\theta)(d+x)=\frac{3267}{9408}m$

 

[Eclair_de_XII]

So I'm just going to attempt part (b) now.

$W=(mg)(sin\theta)(d)=\frac{1}{2}mv^2$
$v=\sqrt{2(mg)(sin\theta)(d)}=\sqrt{2(12kg)(\frac{98}{10}\frac{m}{s^2})(sin\frac{\pi}{6})(\frac{606,656}{940,800}m)}=1.77\frac{m}{s}$

Doing this with $v_y^2=v_0^2+2as$ gives me the same equation. I don't know why I'm off by a factor of -0.1.

 

[haruspex]

So I'm just going to attempt part (b) now.

$W=(mg)(sin\theta)(d)=\frac{1}{2}mv^2$
$v=\sqrt{2(mg)(sin\theta)(d)}=\sqrt{2(12kg)(\frac{98}{10}\frac{m}{s^2})(sin\frac{\pi}{6})(\frac{606,656}{940,800}m)}=1.77\frac{m}{s}$

Doing this with $v_y^2=v_0^2+2as$ gives me the same equation. I don't know why I'm off by a factor of -0.1.

Your expression for v is dimensionally wrong. You forgot to cancel an m.

 

[Eclair_de_XII]

You forgot to cancel an m.

Right, it's: $v=\sqrt{2(g)(sin\theta)(d)}$. But I didn't really take the mass calculation into account when I evaluated the expression. So I'm still off by 0.1, according to the book.

 

[haruspex]

Right, it's: $v=\sqrt{2(g)(sin\theta)(d)}$. But I didn't really take the mass calculation into account when I evaluated the expression. So I'm still off by 0.1, according to the book.

I suspect you have somewhere used an incorrect value from your original (d instead of d+x) calculation.
Please post your complete calculation afresh, showing all steps.

 

[Eclair_de_XII]

Okay.

$U_{spring}=\frac{1}{2}kx^2$
$W=F⋅(d+x)=(mg)(sin\theta)(d+x)$
$(mg)(sin\theta)(d+x)=\frac{1}{2}kx^2$
$d+x=\frac{csc\theta}{2mg}kx^2=\frac{2(10s^2)}{2(12kg)(98m)}(13500\frac{N}{m})(\frac{121}{40000}m^2)=\frac{5s^2}{(6kg)(49m)}(\frac{27}{80})(121J)=\frac{5s^2}{294m⋅kg}(\frac{3267}{80}J)=(\frac{3267}{(16)(294)}m)=\frac{3267}{4704}m$
$d=(d+x)-x=(\frac{3267}{4704}m)-(\frac{11}{200}m)=\frac{1}{940,800}(3267⋅200-11⋅4704)=\frac{1}{940,800}(653,400-51,744)m=\frac{601,656}{940,800}m$

So apparently I made a mistake in the numerator; it's $601,656$, not $606,656$.

$\frac{1}{2}mv^2=(mg)(sin\theta)(d)$
$v^2=2(g)(sin\theta)(d)$
$v=\sqrt{2(g)(sin\theta)(d)}=\sqrt{2(\frac{98}{10}\frac{m}{s^2})(sin\frac{\pi}{6})(\frac{601,656}{940,800}m)}$

I also have to point out that in my previous calculation, I neglected to cancel out $sin\theta$ and $2$; so I actually took the square root of half the value of $(g)(d)$.

 

[haruspex]

Your d+x seems to about double what it should be. The expression with the 13500 looks right, but not the next one.
Instead of all that manual cancellation, it is safer just to plug all the numbers into a calculator or spreadsheet.

 

[Eclair_de_XII]

Your d+x seems to about double what it should be.

Oh, I see what the problem is. I took out 2 from the denominator twice in that expression. That's why.

$d+x=\frac{3267}{9408}m$
$d=(d+x)-x=\frac{3267}{9408}m-\frac{11}{200}m=\frac{1}{1881600}(653400-103488)m=\frac{549,912}{1,881,600}m$
$v=\sqrt{2g(sin\theta)(d)}=\sqrt{2(\frac{98}{10}\frac{m}{s^2})(sin\frac{\pi}{6})(\frac{549,912}{1,881,600}m)}=1.69\frac{m}{s}$

Thank you for your help, again.