1. System Parameters and Strategy The inclined plane has an angle $\theta$ such that $\sin\theta = H/L = 3/5 = 0.6$ and $\cos\theta = \sqrt{1-\sin^2\theta} = 4/5 = 0.8$. The problem requires finding the total time for the object to return to the bottom, starting from the moment the force $F$ is removed. This time is the sum of two parts:

1. $t_{up}$: The time to travel from the point of force removal ($s_0=2$ m) to the highest point on the incline.
2. $t_{down}$: The time to slide from the highest point back to the bottom.

First, we must find the object's velocity, $v_1$, at the moment the force is removed.

2. Phase 1: Motion with Applied Force We define the coordinate system with the x-axis pointing up the incline. While the horizontal force $F$ is applied, the normal force $N_1$ is:

$$N_1 = mg\cos\theta + F\sin\theta$$

The net force along the incline produces acceleration $a_1$:

$$ma_1 = F\cos\theta - mg\sin\theta - f_{k1}$$

where the kinetic friction is $f_{k1} = \mu N_1 = \mu(mg\cos\theta + F\sin\theta)$. The acceleration is:

$$a_1 = \frac{F\cos\theta - mg\sin\theta - \mu(mg\cos\theta + F\sin\theta)}{m}$$

Substituting values:

$a_1 = \frac{100(0.8) - 5(10)(0.6) - 0.3(5(10)(0.8) + 100(0.6))}{5} = \frac{80 - 30 - 0.3(40 + 60)}{5} = \frac{50 - 30}{5} = 4.0 \text{ m/s}^2$

The velocity $v_1$ after traveling a distance $s_0$ from rest is:

$$v_1 = \sqrt{2a_1s_0} = \sqrt{2(4.0)(2)} = 4.0 \text{ m/s}$$

3. Phase 2: Upward Motion After Force Removal With force $F$ removed, the normal force becomes $N_2 = mg\cos\theta$. The object decelerates with $a_2$ due to gravity and friction, both acting down the incline.

$$ma_2 = -mg\sin\theta - \mu N_2 = -mg\sin\theta - \mu mg\cos\theta$$ $$a_2 = -g(\sin\theta + \mu\cos\theta)$$ $a_2 = -10(0.6 + 0.3 \cdot 0.8) = -10(0.84) = -8.4 \text{ m/s}^2$

The time $t_{up}$ to reach the highest point (where velocity is zero) is:

$$t_{up} = \frac{0 - v_1}{a_2} = \frac{-4.0}{-8.4} = \frac{10}{21} \text{ s}$$

4. Phase 3: Downward Motion to the Bottom To find the time to slide down, we first need the total distance from the highest point to the bottom. This is $s_{down} = s_0 + s_{up}$, where $s_{up}$ is the distance traveled during Phase 2.

$$s_{up} = \frac{0^2 - v_1^2}{2a_2} = \frac{-(4.0)^2}{2(-8.4)} = \frac{16}{16.8} = \frac{20}{21} \text{ m}$$

So, the total distance to slide down is $s_{down} = 2 + \frac{20}{21} = \frac{62}{21}$ m. During downward motion, the gravitational component pulls the object down while friction opposes it (acts up). The acceleration $a_3$ is:

$$ma_3 = mg\sin\theta - \mu mg\cos\theta$$ $$a_3 = g(\sin\theta - \mu\cos\theta)$$ $a_3 = 10(0.6 - 0.3 \cdot 0.8) = 10(0.36) = 3.6 \text{ m/s}^2$

Since $a_3 > 0$, the object slides down. The time $t_{down}$ to travel $s_{down}$ from rest is:

$$t_{down} = \sqrt{\frac{2s_{down}}{a_3}} = \sqrt{\frac{2(62/21)}{3.6}} = \sqrt{\frac{310}{189}} \text{ s}$$

5. Total Time The total time to return to the bottom after the force is removed is $T = t_{up} + t_{down}$.

$$T = \frac{10}{21} + \sqrt{\frac{310}{189}} \approx 0.476 + 1.281 \approx 1.757 \text{ s}$$