# Motion of Block Down an Inclined Plane with Friction

• danago

#### danago

Gold Member
http://img297.imageshack.us/img297/1518/88977091un0.gif [Broken]

I managed to derive a solution, however, it doesn't 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:

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

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.

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

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 which the block will move.

Since energy is conserved within the system:

$$W_{friction} = \Delta T + \Delta V_g + \Delta V_e$$

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

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

Anyone see if I am doing anything wrong?

Dan.

Last edited by a moderator: