The rear side of a truck is open and a box of mass `20kg` is placed on the truck `4m` away from the open end `mu = 0.15` and `g = 10m//s^(2)` The truck starts from rest with an acceleration of `2m//s^(2)` on a straight road The box will fall off the truck when it is at a distance from the starting point equal to .

Options

A. `14m`
B. `8m`
C. `16m`
D. `4m`

Explanation

To solve the problem step by step, we will analyze the situation involving the truck and the box placed on it.

### Step 1: Identify the forces acting on the box
The box experiences a pseudo force due to the truck's acceleration. The pseudo force \( F_{\text{pseudo}} \) acting on the box is given by:
\[ F_{\text{pseudo}} = m \cdot a = 20 \, \text{kg} \cdot 2 \, \text{m/s}^2 = 40 \, \text{N} \]

### Step 2: Calculate the maximum static friction force
The maximum static friction force \( F_{\text{friction}} \) that can act on the box is given by:
\[ F_{\text{friction}} = \mu_s \cdot N \]
where \( N \) (the normal force) is equal to the weight of the box:
\[ N = m \cdot g = 20 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 200 \, \text{N} \]
Thus, the maximum static friction force is:
\[ F_{\text{friction}} = 0.15 \cdot 200 \, \text{N} = 30 \, \text{N} \]

### Step 3: Determine the net force acting on the box
The net force \( F_{\text{net}} \) acting on the box is the difference between the pseudo force and the maximum static friction force:
\[ F_{\text{net}} = F_{\text{pseudo}} - F_{\text{friction}} = 40 \, \text{N} - 30 \, \text{N} = 10 \, \text{N} \]

### Step 4: Calculate the acceleration of the box
Using Newton's second law, the acceleration \( a_{\text{box}} \) of the box can be calculated as:
\[ a_{\text{box}} = \frac{F_{\text{net}}}{m} = \frac{10 \, \text{N}}{20 \, \text{kg}} = 0.5 \, \text{m/s}^2 \]

### Step 5: Calculate the time taken for the box to fall off the truck
The box is initially 4 meters from the edge of the truck. Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s = 4 \, \text{m} \), \( u = 0 \), and \( a = 0.5 \, \text{m/s}^2 \):
\[ 4 = 0 + \frac{1}{2} \cdot 0.5 \cdot t^2 \]
This simplifies to:
\[ 4 = 0.25 t^2 \implies t^2 = 16 \implies t = 4 \, \text{s} \]

### Step 6: Calculate the distance traveled by the truck in that time
Using the equation of motion for the truck:
\[ s_{\text{truck}} = ut + \frac{1}{2} a t^2 \]
where \( u = 0 \), \( a = 2 \, \text{m/s}^2 \), and \( t = 4 \, \text{s} \):
\[ s_{\text{truck}} = 0 + \frac{1}{2} \cdot 2 \cdot (4)^2 = \frac{1}{2} \cdot 2 \cdot 16 = 16 \, \text{m} \]

### Conclusion
The truck will have traveled a distance of **16 meters** from its starting point when the box falls off.