To solve the problem step by step, we will analyze the forces acting on the box and the truck's motion.
### Step 1: Understand the scenario
We have a truck accelerating at \(2 \, \text{m/s}^2\) with a box placed \(8 \, \text{m}\) from the open end. The box has a mass of \(2 \, \text{kg}\) and the coefficient of friction (\(\mu\)) between the box and the truck is \(0.1\).
### Step 2: Calculate the normal force
The normal force (\(N\)) acting on the box is equal to its weight since there is no vertical motion:
\[
N = mg = 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N}
\]
### Step 3: Calculate the maximum frictional force
The maximum frictional force (\(f_{\text{max}}\)) that can act on the box is given by:
\[
f_{\text{max}} = \mu N = 0.1 \times 20 \, \text{N} = 2 \, \text{N}
\]
### Step 4: Calculate the maximum possible acceleration of the box
Using Newton's second law, the maximum acceleration (\(a_{\text{max}}\)) that the box can achieve due to friction is:
\[
f_{\text{max}} = ma_{\text{max}} \implies a_{\text{max}} = \frac{f_{\text{max}}}{m} = \frac{2 \, \text{N}}{2 \, \text{kg}} = 1 \, \text{m/s}^2
\]
### Step 5: Determine relative motion
The truck accelerates at \(2 \, \text{m/s}^2\) while the box can only accelerate at \(1 \, \text{m/s}^2\). Thus, the box will experience relative motion with respect to the truck.
### Step 6: Calculate the relative acceleration
The relative acceleration of the box with respect to the truck is:
\[
a_{\text{relative}} = a_{\text{truck}} - a_{\text{box}} = 2 \, \text{m/s}^2 - 1 \, \text{m/s}^2 = 1 \, \text{m/s}^2
\]
### Step 7: Use kinematics to find the time until the box falls off
We can use the second equation of motion to find the time (\(t\)) it takes for the box to travel \(8 \, \text{m}\) relative to the truck:
\[
s = ut + \frac{1}{2} a_{\text{relative}} t^2
\]
Here, \(s = 8 \, \text{m}\), \(u = 0\), and \(a_{\text{relative}} = 1 \, \text{m/s}^2\):
\[
8 = 0 + \frac{1}{2} \times 1 \times t^2 \implies 8 = \frac{1}{2} t^2 \implies t^2 = 16 \implies t = 4 \, \text{s}
\]
### Step 8: Calculate the distance traveled by the truck in that time
Now, we calculate the distance traveled by the truck in \(4 \, \text{s}\):
\[
d = ut + \frac{1}{2} a t^2
\]
Here, \(u = 0\), \(a = 2 \, \text{m/s}^2\):
\[
d = 0 + \frac{1}{2} \times 2 \times (4^2) = 0 + \frac{1}{2} \times 2 \times 16 = 16 \, \text{m}
\]
### Final Answer
The box will fall off the truck when it is at a distance from the starting point equal to:
\[
\text{Distance from starting point} = 16 \, \text{m}
\]
