## Physics Friday 149

For a block sliding down an inclined surface with angle of incline θ and with coefficient of kinetic friction μ. If we consider the forces, we see that the block can slide down the incline with constant velocity when the net forces are zero. Balancing forces perpendicular to the plane, the normal force is thus $N=mg\cos\theta$. The kinetic friction is thus $f=\mu{N}=\mu{mg}\cos\theta$, while the component of gravity parallel to the plane is thus $mg\sin\theta$. We see that these cancel when
$\mu\cos\theta=\sin\theta$, or when $\mu=\tan\theta$.

Now, suppose we have a large incline with $\mu=\tan\theta$, and we start our block sliding with a velocity v0 in a horizontal direction; that is to say, along the plane in a direction perpendicular to the direction of the slope. What, then, will be the speed a long time later?

Let v be the total speed of the block, and let vy be the velocity component in the downslope direction. Now, we have two forces parallel to the plane; the friction force, with magnitude
$f=\mu{N}=\mu{mg}\cos\theta=mg\sin\theta$
directed opposite to the motion of the block, and the component of gravity, also of magnitude $mg\sin\theta$, directed downslope.
The former produces an acceleration of magnitude $g\sin\theta$ opposite the direction of motion, while the gravity gives the same acceleration $g\sin\theta$ in the downslope direction; thus
$\frac{dv}{dt}=-g\sin\theta$, and $\frac{dv_y}{dt}=g\sin\theta$, so that
$\frac{dv}{dt}+\frac{dv_y}{dt}=0$, and thus
$v+v_y=C$, for C some constant.
Now, at time t=0, our motion is purely horizontal, and so vy=0, v=v0, and so
C=v0.
Now, after a long time, the horizontal component of the velocity will be effectively zero, so that the velocity is entirely downslope. Then v=vy, and so $v+v_y=v_0$ becomes
$2v=v_0$,
and so the speed after a long time is
$v=v_y=\frac12v_0$, half the initial speed; now in the downslope direction.