Answer
(a) The box slides a distance of 0.87 meters up the slope.
(b) The total time to return to the starting point is 1.33 seconds.
Work Step by Step
Note that when the box is sliding up, the force of kinetic friction is directed down the slope because it opposes the motion of the box. Similarly, when the box is sliding down, the force of kinetic friction is directed up the slope.
(a) We can use a force equation to find the magnitude of the acceleration.
$ma = \sum F$
$ma = mg~sin(\theta) + mg~cos(\theta)\cdot \mu_k$
$a = g~sin(\theta) + g~cos(\theta)\cdot \mu_k$
$a = (9.80~m/s^2)~sin(25.0^{\circ}) + (9.80~m/s^2)~cos(25.0^{\circ})(0.12)$
$a = 5.2~m/s^2$
As the box slides up the slope, the magnitude of deceleration is $5.2~m/s^2$.
$x = \frac{v-v_0^2}{2a} = \frac{0-(3.0~m/s)^2}{(2)(-5.2~m/s^2)}$
$x = 0.87~m$
The box slides a distance of 0.87 meters up the slope.
(b) Let $t_1$ be the time moving up the slope and let $t_2$ be the time moving down the slope.
$t_1 = \frac{v-v_0}{a} = \frac{0-3.0~m/s}{-5.2~m/s^2} = 0.58~s$
Before we find $t_2$, we need to find the acceleration when the box is sliding down. We can use a force equation to find the acceleration.
$ma = \sum F$
$ma = mg~sin(\theta) - mg~cos(\theta)\cdot \mu_k$
$a = g~sin(\theta) - g~cos(\theta)\cdot \mu_k$
$a = (9.80~m/s^2)~sin(25.0^{\circ}) - (9.80~m/s^2)~cos(25.0^{\circ})(0.12)$
$a = 3.1~m/s^2$
We can use this acceleration to find $t_2$.
$x = \frac{1}{2}at_2^2$
$t_2 = \sqrt{\frac{2x}{a}} = \sqrt{\frac{(2)(0.87~m)}{3.1~m/s^2}}$
$t_2 = 0.75~s$
The total time to return to the starting point is $t_1+t_2$ which is 1.33 seconds.
