A 40 kg box is at rest 5 m from the end of a truck bed. The coefficient of friction between the box and the truck bed is 0.15. If the truck accelerates at 2 m/s2, how long will it take for the box to

Mass of the box, $m = 40 \text{ kg}$
Initial distance of the box from the open end of the truck, $d = 5 \text{ m}$
Coefficient of friction between the box and the truck surface, $\mu = 0.15$
Acceleration of the truck, $a_T = 2 \text{ m/s}^2$
We will use the acceleration due to gravity, $g = 10 \text{ m/s}^2$.

The normal force ($N$) acting on the box is equal to its weight, $mg$.

The maximum static friction force ($f_{friction}$) is given by $\mu N$. Once the box starts slipping, this becomes the kinetic friction force.

The friction force is the only horizontal force acting on the box. Using Newton's second law ($F = ma$), the acceleration of the box ($a_{box}$) is:

The truck accelerates at $a_T = 2 \text{ m/s}^2$, and the box accelerates at $a_{box} = 1.5 \text{ m/s}^2$. Since $a_T > a_{box}$, the truck moves forward faster than the box, causing the box to slip backward relative to the truck.
The relative acceleration ($a_{rel}$) of the truck with respect to the box (or the box slipping backward relative to the truck) is:

The box falls off when it has moved a relative distance of $d = 5 \text{ m}$ with respect to the truck. Both the truck and the box start from rest, so the initial relative velocity is 0.
Using the kinematic equation $s = ut + \frac{1}{2}at^2$:

The question asks for the distance from the starting point where the box falls off. This is the distance traveled by the truck during the time $t$.
Using the kinematic equation for the truck: $S_T = U_T t + \frac{1}{2}a_T t^2$.
Since the truck starts from rest, $U_T = 0$.

The distance from the starting point where the box falls off the truck is 20 meters. This corresponds to option (a).

The final answer is $\text{(a)}$.