A wooden box is placed on the back part of a lorry moving with an acceleration of `6ms^(-2)`, if `mu = 0.5`, the acceleration of the box relative to lorry is

Options

A. `1.1 ms^(-2)`
B. `2 ms^(-2)`
C. `1 . 5 ms^(-2)`
D. `0`

Answer

To solve the problem of finding the acceleration of the wooden box relative to the lorry, we can follow these steps:

### Step 1: Identify the forces acting on the box
The box experiences the following forces:
- The gravitational force acting downward, \( mg \)
- The normal force acting upward, \( N \)
- The frictional force acting horizontally, which prevents the box from sliding off the lorry.

### Step 2: Determine the normal force
Since the box is on a horizontal surface, the normal force \( N \) is equal to the weight of the box:
\[ N = mg \]

### Step 3: Calculate the maximum frictional force
The maximum static frictional force \( f_{\text{max}} \) can be calculated using the coefficient of friction \( \mu \):
\[ f_{\text{max}} = \mu N = \mu mg \]
Given \( \mu = 0.5 \):
\[ f_{\text{max}} = 0.5 mg \]

### Step 4: Determine the acceleration of the lorry
The lorry is moving with an acceleration \( a_{\text{lorry}} = 6 \, \text{m/s}^2 \).

### Step 5: Calculate the maximum possible acceleration of the box
The maximum possible acceleration \( a_{\text{box}} \) of the box due to friction can be calculated as:
\[ a_{\text{box}} = \frac{f_{\text{max}}}{m} = \frac{0.5 mg}{m} = 0.5g \]
Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ a_{\text{box}} = 0.5 \times 9.8 = 4.9 \, \text{m/s}^2 \]

### Step 6: Compare the accelerations
The box can accelerate at a maximum of \( 4.9 \, \text{m/s}^2 \) due to friction, while the lorry accelerates at \( 6 \, \text{m/s}^2 \). Since \( 4.9 \, \text{m/s}^2 < 6 \, \text{m/s}^2 \), the box will slide backward relative to the lorry.

### Step 7: Calculate the relative acceleration
The relative acceleration \( a_{\text{relative}} \) of the box with respect to the lorry is given by:
\[ a_{\text{relative}} = a_{\text{lorry}} - a_{\text{box}} \]
Substituting the values:
\[ a_{\text{relative}} = 6 - 4.9 = 1.1 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the box relative to the lorry is \( 1.1 \, \text{m/s}^2 \).

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