1. Acceleration during Upward Motion

We analyze the forces acting on the object as it moves up the incline. The net force is directed down the slope, opposing the motion. The forces along the incline are the component of gravity, $mg\sin\alpha$, and the kinetic friction force, $f_k = \mu N$. The normal force is $N = mg\cos\alpha$.

Applying Newton's second law along the incline (taking the upward direction as positive):

$$ \sum F_{up} = -mg\sin\alpha - \mu mg\cos\alpha = m a_{up} $$

The acceleration (deceleration) during the upward motion is:

$$ a_{up} = -g(\sin\alpha + \mu\cos\alpha) $$

Let the magnitude of this deceleration be $a_1 = g(\sin\alpha + \mu\cos\alpha)$.

2. Acceleration during Downward Motion

As the object slides down, the kinetic friction force reverses direction and points up the incline. Applying Newton's second law along the incline (taking the downward direction as positive):

$$ \sum F_{down} = mg\sin\alpha - \mu mg\cos\alpha = m a_{down} $$

The acceleration during the downward motion is:

$$ a_{down} = g(\sin\alpha - \mu\cos\alpha) $$

Let this acceleration be $a_2 = g(\sin\alpha - \mu\cos\alpha)$.

3. Kinematic Relation for Distance

Let $d$ be the distance traveled up the slope. We can express this distance using the time of travel for each segment.

For the upward trip, the object starts with an initial velocity and comes to rest after time $t_1$. Using the kinematic equation $d = \frac{1}{2}at^2$ is simplest if we consider the motion in reverse: an object starting from rest and accelerating with $a_1$ over time $t_1$. The distance covered is:

$$ d = \frac{1}{2} a_1 t_1^2 $$

For the downward trip, the object starts from rest and accelerates with $a_2$ over time $t_2$. The distance covered is:

$$ d = \frac{1}{2} a_2 t_2^2 $$

4. Deriving the Ratio

The distance $d$ is the same for both the upward and downward journeys. By equating the two expressions for $d$:

$$ \frac{1}{2} a_1 t_1^2 = \frac{1}{2} a_2 t_2^2 $$ $$ a_1 t_1^2 = a_2 t_2^2 $$

We can now solve for the ratio $t_2/t_1$:

$$ \frac{t_2^2}{t_1^2} = \frac{a_1}{a_2} $$

Substituting the expressions for $a_1$ and $a_2$:

$$ \frac{t_2^2}{t_1^2} = \frac{g(\sin\alpha + \mu\cos\alpha)}{g(\sin\alpha - \mu\cos\alpha)} = \frac{\sin\alpha + \mu\cos\alpha}{\sin\alpha - \mu\cos\alpha} $$

Taking the square root of both sides gives the final result:

$$ \frac{t_2}{t_1} = \sqrt{\frac{\sin\alpha + \mu\cos\alpha}{\sin\alpha - \mu\cos\alpha}} $$