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. 4m
B. 8m
C. 16m
D. 32m

Answer

To solve the problem step by step, we will analyze the forces acting on the box and use the equations of motion to determine the distance the truck travels when the box falls off.

### Step 1: Identify the forces acting on the box
The box experiences:
- Gravitational force (weight) downward: \( F_g = mg \)
- Normal force \( N \) upward
- Frictional force \( F_f \) opposing the motion of the truck: \( F_f = \mu N \)
- Pseudo force due to the truck's acceleration acting on the box in the opposite direction of the truck's acceleration.

### Step 2: Calculate the normal force
Since the box is on a horizontal surface, the normal force \( N \) is equal to the weight of the box:
\[ N = mg = 20 \, \text{kg} \times 10 \, \text{m/s}^2 = 200 \, \text{N} \]

### Step 3: Calculate the frictional force
Using the coefficient of friction:
\[ F_f = \mu N = 0.15 \times 200 \, \text{N} = 30 \, \text{N} \]

### Step 4: Determine the effective acceleration of the box
The truck accelerates at \( a = 2 \, \text{m/s}^2 \). The box will experience a pseudo force due to this acceleration:
\[ F_{\text{pseudo}} = ma = 20 \, \text{kg} \times 2 \, \text{m/s}^2 = 40 \, \text{N} \]

The net force acting on the box can be expressed as:
\[ F_{\text{net}} = F_{\text{pseudo}} - F_f \]
\[ F_{\text{net}} = 40 \, \text{N} - 30 \, \text{N} = 10 \, \text{N} \]

### Step 5: Calculate the acceleration of the box
Using Newton's second law \( F = ma \):
\[ F_{\text{net}} = ma' \]
\[ 10 \, \text{N} = 20 \, \text{kg} \cdot a' \]
\[ a' = \frac{10 \, \text{N}}{20 \, \text{kg}} = 0.5 \, \text{m/s}^2 \]

### Step 6: Calculate the time taken for the box to fall off
The box is initially \( 4 \, \text{m} \) from the edge of the truck. We can use the equation of motion:
\[ d = ut + \frac{1}{2} a' t^2 \]
Since the initial velocity \( u = 0 \):
\[ 4 \, \text{m} = 0 + \frac{1}{2} \times 0.5 \, \text{m/s}^2 \times t^2 \]
\[ 4 = 0.25 t^2 \]
\[ t^2 = \frac{4}{0.25} = 16 \]
\[ t = 4 \, \text{s} \]

### Step 7: Calculate the distance traveled by the truck in that time
Using the truck's acceleration:
\[ d_{\text{truck}} = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \):
\[ d_{\text{truck}} = 0 + \frac{1}{2} \times 2 \, \text{m/s}^2 \times (4 \, \text{s})^2 \]
\[ d_{\text{truck}} = 1 \times 16 = 16 \, \text{m} \]

### Final Answer
The box will fall off the truck when it is at a distance from the starting point equal to **16 meters**.

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