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Sliding Block

  • Thread starter danago
  • Start date
  • #1
danago
Gold Member
1,122
4
http://img297.imageshack.us/img297/1518/88977091un0.gif [Broken]

I managed to derive a solution, however, it doesnt seem to give me the correct answer. Here is what i did:

I first defined my positive x axis to go down the slope and my positive y axis to be upward and normal to the slope. The forces acting on the block before it hits the spring are:

[tex]
\begin{array}{l}
\overrightarrow W = \left( {\begin{array}{*{20}c}
{mg\sin \theta } \\
{ - mg\cos \theta } \\
\end{array}} \right) \\
\overrightarrow N = \left( {\begin{array}{*{20}c}
0 \\
{mg\cos \theta } \\
\end{array}} \right) \\
\overrightarrow F = - \mu _k \left| {\overrightarrow N } \right|\frac{{\overrightarrow v }}{{\left| {\overrightarrow v } \right|}} = \left( {\begin{array}{*{20}c}
{ - \mu _k mg\cos \theta } \\
0 \\
\end{array}} \right) \\
\end{array}
[/tex]

Where W is the weight, F is the kinetic friction and N is the normal force.

As it moves down the slope and compresses the spring, it will lost gravitational potential energy, gain elastic potential energy, lose kinetic energy and lose energy due to friction.

[tex]
\begin{array}{l}
\Delta V_g = mg\Delta h = - mg(\Delta x + \delta )\sin \theta \\
\Delta V_e = \frac{1}{2}k\Delta (x^2 ) = \frac{1}{2}k(\Delta x)^2 \\
\Delta T = - \frac{1}{2}mv^2 \\
W_{friction} = - \mu _k mg(\delta + \Delta x)\cos \theta \\
\end{array}
[/tex]

Where Vg/e are the gravitational/elastic potential energies, T is the kinetic energy and Wfriction is the work done by the friction on the block i.e. energy lost due to friction. Delta x is the compression of the spring, so delta x + delta is the total distance down the slope wich the block will move.

Since energy is conserved within the system:

[tex]
W_{friction} = \Delta T + \Delta V_g + \Delta V_e
[/tex]

Substituting my expressions for the different energies into this gives me a quadratic equation in delta x, which has the following solution:

[tex]
\begin{array}{l}
\Delta x = \frac{{ - B \pm \sqrt {B^2 - 4AC} }}{{2A}} \\
\\
A = 0.5k \\
B = mg(\mu _k \cos \theta - \sin \theta ) \\
C = mg\delta (\mu _k \cos \theta - \sin \theta ) - 0.5mv^2 \\
\end{array}
[/tex]

Anyone see if im doing anything wrong?

Thanks in advance,
Dan.
 
Last edited by a moderator:

Answers and Replies

  • #2
danago
Gold Member
1,122
4
Plugging my values into my solution i get -0.256 and 0.286. I would think that id take the latter as my solution since ive treated delta x as a positive quantity through out the derivation of my solution.
 
  • #3
alphysicist
Homework Helper
2,238
1
Hi danago,

That looks right to me; the positive solution would be the answer you wanted for this problem.
 
  • #4
danago
Gold Member
1,122
4
Thanks for the reply :smile: The working just got a little messy so i thought perhaps i had made an algebraic error, but i guess the solution given to me could always be at fault.
 

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