1. **Problem statement:** A 20 kg cube is placed 5 m from the rear of a truck on a horizontal floor. The coefficient of friction between the cube and the floor is 0.1. The truck accelerates from rest at $2\ \text{m/s}^2$. We need to find how long it takes for the cube to fall off the back of the truck. 2. **Relevant formulas and concepts:** - Frictional force $f = \mu mg$ where $\mu$ is the coefficient of friction, $m$ is mass, and $g$ is acceleration due to gravity ($9.8\ \text{m/s}^2$). - Maximum acceleration the cube can have without slipping is $a_{max} = \mu g$. - If the truck's acceleration $a$ is greater than $a_{max}$, the cube will slip. - The cube will start slipping and move relative to the truck floor. 3. **Calculate maximum frictional acceleration:** $$a_{max} = \mu g = 0.1 \times 9.8 = 0.98\ \text{m/s}^2$$ 4. **Compare truck acceleration with $a_{max}$:** Truck acceleration $a = 2\ \text{m/s}^2 > a_{max} = 0.98\ \text{m/s}^2$ This means the cube cannot accelerate as fast as the truck and will slip backward relative to the truck. 5. **Relative acceleration of cube with respect to truck:** $$a_{relative} = a - a_{max} = 2 - 0.98 = 1.02\ \text{m/s}^2$$ 6. **Calculate time for cube to slip 5 m relative to truck:** Using equation for displacement with initial velocity zero: $$s = \frac{1}{2} a_{relative} t^2$$ $$5 = \frac{1}{2} \times 1.02 \times t^2$$ $$t^2 = \frac{5 \times 2}{1.02} = \frac{10}{1.02} \approx 9.8$$ $$t = \sqrt{9.8} \approx 3.13\ \text{seconds}$$ **Final answer:** The cube will fall off the back of the truck after approximately **3.13 seconds**.