Question

A crate is given a quick push at the bottom of a loading ramp. The box slides up the ramp, and then slides back down. If there is friction with the ramp (as there always is) which takes longer, the slide up the ramp or the slide back down?

Answer 1

Answer:

The slide back down takes longer than the slide up.

Explanation:

In this is problem, the loss of energy is a key factor.

When the box is pushed up the ramp, it gains a specific speed, and an associated kinetic energy. As it slides up, it losses kinetic energy due to the gravitational and frictional forces, and it is transformed to gravitational potential energy (or potential energy). When it reaches its highest point, the speed becomes cero and all its kinetic energy is converted to potential energy.

However, since there is a frictional force, some amount of energy is loss, so the potential energy is lesser than the initial kinetic energy.

The same occurs when the box slides back down: its potential energy is converted into kinetic energy, but some energy is loss due to the frictional force. This means that the final kinetic energy is smaller than the initial, which implies that its final velocity is smaller than the initial velocity.

Using the acceleration definition:

$a=\frac{v_f-v_o}{t}$

We solve for the time (t) and get:

$t=\frac{v_f-v_o}{a}$

We have to use this equation two times, one for each movement  (slide up and slide down).

The gravitational force points down with the same magnitude in both, but the frictional force goes down when the box is sliding up and it goes up when the box slides down. So the magnitude of the acceleration when it goes up is smaller than the other one.

If we analyze both situations:

A) When it goes up:

$t_{up}=-\frac{v_o}{a_1}$

B) When it goes down:

$t_{down}=\frac{v_f}{a_2}$

And we have that $v_o>v_f$ and $a_1$. Which implies that, in any case, $t_{up}>t_{down}$.