# Find the time at which a particle projected up an inclined plane, comes to rest?

A particle of mass $m$ is projected up a plane that is inclined at an angle $\alpha$ to the horizontal. At $t=0$, its velocity is $v_0$ and the coefficient of dynamic friction of the slope is $\mu$. Show that the particle comes to rest at time $T = \dfrac{v_0}{g(\mu \cos\alpha + > \sin\alpha)}$

I remember doing these kinds of questions at A level but I'm pretty stuck here. I drew a diagram and considered the forces acting on the particle:

$m\vec{g} = -mg\sin\alpha\hat{i} - mg\sin\alpha\hat{j}$ (Taking the upwards direction to be positive)

$\vec{v_0} = v_{0}\sin\alpha\hat{i} + v_{0}\cos\alpha\hat{j}$

$\vec{R} = R\hat{j} = \mu mg\cos\alpha$

$\vec{F} = \mu\vec{R}$

That's about as far as I can get at this point!

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Double check your expression for $m\vec g$. The normal force is $\mu mg\cos \alpha$. Also, $v_0$ is the initial velocity in the answer. You have it as the velocity function. – John Douma Apr 25 '13 at 19:47
I'm not sure how to go about finding $T$ given the forces... – Mathlete Apr 25 '13 at 19:50
Draw the incline plane. This is a one dimensional problem. The angle $\alpha$ is the angle between the weight vector and the normal vector. The $\sin \alpha$ is the component along the plane and the $\cos \alpha$ is the component perpendicular to the plane. – John Douma Apr 25 '13 at 19:51
Yep I've done that, I think the forces are all now correct? – Mathlete Apr 25 '13 at 19:52
You can divide to get rid of the mass and you end up with $\vec g=\ddot{x} = -g\sin \alpha - \mu g\cos \alpha$. Integrate that to get the velocity. Then $v_0$ will be the initial velocity. Set $t=0$ and solve. – John Douma Apr 25 '13 at 19:57

You have correctly set up your forces as $$ma = -mg\sin \alpha -\mu mg \cos \alpha$$.

We use the fact that $a=\ddot{x}$ and divide by $m$ to get $$\ddot{x}=-g\sin \alpha - \mu g\cos \alpha$$.

Integrating with respect to $t$ we get

$$v = \dot{x} = (-g\sin \alpha - \mu g\cos \alpha)t + v_0$$

Finally, setting $v=0$, we get

$$0=(-g\sin \alpha - \mu g\cos \alpha)t + v_0\Rightarrow t= \frac{v_0}{g\sin \alpha + \mu g\cos \alpha}$$

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How would I incorporate the $\hat{i}$ and $\hat{j}$ components? – Mathlete Apr 25 '13 at 20:15
You don't. This is a one dimensional problem. Even though the normal force is in the $\bar j$ direction, the frictional force is not. – John Douma Apr 25 '13 at 20:17
OK...I just feel like you've skipped a few steps in the working out. – Mathlete Apr 25 '13 at 20:20
Please be more specific. I am happy to clarify anything. – John Douma Apr 25 '13 at 20:21
The first step, how did you get the $\mu$? – Mathlete Apr 25 '13 at 20:23