Question

c. state what happens to the kinetic energy of the crate as it slides along the incline.

Explanation:

When a crate slides along an incline (assuming it's sliding down and there's no mention of external forces opposing motion to a great extent, or if sliding up against gravity but typically we consider sliding down first), if we consider a scenario with friction, the kinetic energy's change depends on forces. But in a common case, if the crate is sliding down a frictionless incline, its speed increases as it falls (loses gravitational potential energy, gains kinetic energy). If there is friction, the kinetic energy will still increase but less than frictionless, or if sliding up, it would decrease. But generally, when sliding down an incline (most typical case), as the crate moves down, its speed increases (since the component of gravitational force along the incline causes acceleration, assuming net force is down the incline). Kinetic energy is given by $KE = \frac{1}{2}mv^2$, so as speed $v$ increases, kinetic energy increases (assuming mass $m$ is constant). If there were no friction, mechanical energy (KE + PE) would be conserved, so PE (gravitational) decreases and KE increases. If there is friction, some energy is lost to heat, but the KE still increases (because the net force down the incline causes acceleration, so speed increases) until maybe it reaches a terminal velocity (but for a crate sliding on a typical incline, unless friction is very high, speed increases). So the key is: as the crate slides down the incline (assuming the usual case of moving down), its kinetic energy increases because its speed increases (due to the component of gravity along the incline causing acceleration, and $KE$ is proportional to the square of speed).

Answer:

As the crate slides down the incline (assuming it is moving downwards), its kinetic energy increases. This is because the crate’s speed increases (due to the component of gravitational force along the incline causing acceleration, or conversion of gravitational potential energy to kinetic energy, with $KE=\frac{1}{2}mv^2$ and speed $v$ increasing as it moves down), so kinetic energy (proportional to $v^2$) increases. If sliding up, kinetic energy would decrease as speed decreases, but the typical scenario is sliding down, so kinetic energy increases.