A box is placed on the floor of a truck moving with an acceleration of `7 ms^(-2)`. If the coefficient of kinetic friction between the box and surface of the truck is 0.5, find the acceleration of the box relative to the truck

Options

A. `1.7 ms^(-2)`
B. `2.1 ms^(-2)`
C. `3.5 ms^(-2)`
D. `4.5 ms^(-2)`

Answer

To find the acceleration of the box relative to the truck, we can follow these steps:

### Step 1: Identify the forces acting on the box
The box experiences two main forces:
1. The pseudo force due to the truck's acceleration.
2. The frictional force opposing the motion of the box.

### Step 2: Calculate the pseudo force
The pseudo force \( F_{\text{pseudo}} \) acting on the box can be calculated using the formula:
\[ F_{\text{pseudo}} = m \cdot a \]
where \( a \) is the acceleration of the truck. Given that \( a = 7 \, \text{m/s}^2 \), we have:
\[ F_{\text{pseudo}} = m \cdot 7 \]

### Step 3: Calculate the frictional force
The frictional force \( F_{\text{friction}} \) can be calculated using the formula:
\[ F_{\text{friction}} = \mu \cdot m \cdot g \]
where \( \mu \) is the coefficient of kinetic friction and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). Given \( \mu = 0.5 \), we have:
\[ F_{\text{friction}} = 0.5 \cdot m \cdot 9.8 = 4.9m \]

### Step 4: Set up the equation of motion
In the non-inertial frame of the truck, the box will accelerate backwards due to the pseudo force and forwards due to the frictional force. The net force acting on the box can be expressed as:
\[ F_{\text{net}} = F_{\text{pseudo}} - F_{\text{friction}} \]
Substituting the expressions for the forces:
\[ F_{\text{net}} = (7m) - (4.9m) = (7 - 4.9)m = 2.1m \]

### Step 5: Calculate the acceleration of the box relative to the truck
Using Newton's second law, \( F = ma \), we can find the acceleration \( a_{\text{box}} \) of the box relative to the truck:
\[ F_{\text{net}} = m \cdot a_{\text{box}} \]
Thus,
\[ 2.1m = m \cdot a_{\text{box}} \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ a_{\text{box}} = 2.1 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the box relative to the truck is:
\[ \boxed{2.1 \, \text{m/s}^2} \]