# A spring-block system on a frictional plane

Maybe I should have posted this question in 'elementary physics' but the question seemed rather difficult to me(although involving elementary concepts) so i am posting it at this forum.Here goes.

## Homework Statement

A block of mass m is attached to a spring of spring constant k.the other end of the spring is attached to a wall.The mass rests on a surface with friction coefficient $$\mu$$.Now the mass is given a displacement $$x_{0}$$ from its equilibrium position.Calculate x at any subsequent time.

## Homework Equations

The forces acting on the block are(taking the equilibrium position as origin)
force due to the spring:
$$\\F_{spring} = -kxi\hat \\$$

$$\\ F_{friction} = \mu mgi\hat \\ \mbox{ when the block is moving in the -i direction}$$

$$F_{friction} = -\mu mgi\hat \\ \mbox{ when the block is moving in +i direction}$$

## The Attempt at a Solution

The equations i obtain are
$$m\ddot x +kx = \mu mg\ \mbox{when moving towards -i} \\$$
$$\\ \mbox{and} m\ddot x + kx = -\mu mg\ \mbox{when moving towards +i}$$
Solving them i get,for the first half cycle:
$$\\x = (x_{0} - \frac{\mu mg}{k})\cos \frac{k}{m}t + \frac{\mu mg}{k}\\$$
$$\mbox{At the end of the first half cycle,the position of the block is,then } x'_{0} = -x_{o} + 2\frac{\mu mg}{k}$$
$$\\ \mbox{Now,if i take t= 0 at} x= x'_{0} \ \mbox{ ,i get,from solving the second differential equation}$$
$$\\x = -( x'_{o}+ \frac{\mu mg}{k}) \cos \frac{k}{m}t - \frac{\mu mg}{k}$$
$$\mbox{At the end of this half-cycle the position is }$$
$$x_{0} - 4\frac{\mu mg}{k}$$

So i'm getting the result that for each half cycle the distance traversed goes down by $$2\frac{\mu mg}{k}$$.I get the same reult from work-energy theorem.
is my method correct?if yes,is there a more elegant way of doing it?is there at least an elegant way of writing the answer?In my answer i'm putting t=0 at the start of each half-cycle,so i can't boil it down to less than 2 equations.

P.S:this is te first time i'm using latex.

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