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