Inclined planes

The plane is smooth, therefore only the weight and the normal force act on the block. The $x$ axis is chosen along the plane, pointing down, and the $y$ axis perpendicular to the plane. The weight is decomposed into $mg\sin \alpha$ along the plane and $mg\cos \alpha$ perpendicular to the plane; the block does not leave the plane, so along $y$ the acceleration is zero and the normal force balances the perpendicular component. This gives the results of the smooth inclined plane, $a=g\sin \alpha$ and $N=mg\cos \alpha$.

The acceleration is constant and the block starts from rest, therefore the motion along the plane is uniformly accelerated. From the equation of motion one obtains the time to travel the $3.0\mathrm{m}$, and from the relation between velocity and distance the final velocity. As a check, $v=at=4.905\cdot 1.106m/s=5.42m/s$. The velocity at the bottom is the same one would have in free fall from a height $d\sin \alpha =1.5\mathrm{m}$, since $\sqrt{2\cdot 9.81\cdot 1.5}m/s=5.42m/s$; the reason becomes clear in the topic on work and energy.

At constant velocity the acceleration is zero, therefore the net force is zero in both cases. The axes of the inclined plane are used: $x$ along the plane pointing up, $y$ perpendicular. In the first case the applied force has only the component along $x$ and must balance the component of the weight along the plane, $mg\sin \alpha =16.8\mathrm{N}$.

In the second case the horizontal force forms the angle $\alpha$ with the plane, therefore it has a component $F\cos \alpha$ along the plane and a component $F\sin \alpha$ that presses the crate against the plane. The first equation gives $F=mg\tan \alpha =17.9\mathrm{N}$, a value greater than that with the parallel force, because only part of the horizontal force is useful for pushing along the plane. The second equation gives the normal force $52.2\mathrm{N}$, greater than $mg\cos \alpha =46.1\mathrm{N}$ because of the component of $F$ that pushes against the plane.

To determine whether the block slides, the condition of the angle of repose is used, according to which a body at rest remains at rest if and only if $\tan \alpha \le {\mu }_s$. Since $\tan 35^{\circ }=0.700$ exceeds ${\mu }_s=0.40$, static friction cannot balance the component of the weight along the plane, and the block starts to slide.

During sliding the friction is kinetic, has magnitude ${\mu }_{d}N={\mu }_{d}mg\cos \alpha$ and is directed up along the plane, opposite to the motion. The second law along the plane gives the acceleration of the inclined plane with friction, $a=3.62m/{\mathrm{s}}^2$, which does not depend on the mass; the mass is needed only to compute the forces, for example $N=12.1\mathrm{N}$ and $f_d=3.01\mathrm{N}$. The motion is uniformly accelerated from rest, therefore after $2.0\mathrm{m}$ the velocity is $3.80m/s$. As a check, the acceleration is smaller than that on the smooth plane, $g\sin 35^{\circ }=5.63m/{\mathrm{s}}^2$.

The problem divides into two phases, because kinetic friction changes sense when the sense of motion changes. On the way up the block moves upward, therefore friction is directed downward, like the component of the weight along the plane, and the two forces add and the deceleration is $a_s=6.81m/{\mathrm{s}}^2$. The motion is uniformly decelerated, so the block stops after $d=2.64\mathrm{m}$.

At the highest point the block is at rest for an instant, therefore to know whether it slides back down the condition of the angle of repose is used. Since $\tan 25^{\circ }=0.466$ exceeds ${\mu }_s=0.40$, the block slides back down. On the way down friction is directed upward and opposes the component of the weight, so the acceleration is much smaller, $a_d=1.48m/{\mathrm{s}}^2$. Traveling back the $2.64\mathrm{m}$ starting from rest, the block passes the launch point again at $2.80m/s$, less than half the initial velocity, because in both phases friction has taken energy away from the motion.