Answer
(a) x = 2.8 m
(b) t = 2.4 seconds
Work Step by Step
(a) We can use a force equation to find the acceleration:
$ma = F$
$ma = mg ~sin(\theta)$
$a = g ~sin(\theta) = (9.80 ~m/s^2) \cdot ~sin(22.0^{\circ})$
$a = 3.67~m/s^2$
When the block slides up the ramp, the rate of deceleration is $a = 3.67~m/s^2$.
We can use kinematics to find the distance:
$x = \frac{v^2-v_0^2}{2a} = \frac{0 -(4.5 ~m/s)^2}{(2)(-3.67 ~m/s^2)} = 2.8 ~m$
(b) We can use kinematics to find the time t to slide up to the block's highest point on the ramp:
$t = \frac{v-v_0}{a} = \frac{0 - 4.5 ~m/s}{-3.67 ~m/s^2} = 1.2 ~s$
Since we ignore friction, the total time to return to the starting point is 2t which is 2.4 seconds.
