A block is placed at distance of 2m from the rear on the floor of a truck (g=10`ms^(-2)`). When the truck moves with an acceleration of `8ms^(-2)`, the block takes 2 sec to fall off from the rear of the truck. The coefficient of sliding friction between truck and the block is

Options

A. `0.5`
B. `0.1`
C. `0.8`
D. `0.7`

Answer

To solve the problem step by step, we will analyze the motion of the block and the forces acting on it.

### Step 1: Understand the scenario
- A block is placed on the floor of a truck that is accelerating forward with an acceleration \( A_1 = 8 \, \text{m/s}^2 \).
- The block is initially at rest and is located 2 meters from the rear of the truck.
- The block takes 2 seconds to fall off the rear of the truck.

### Step 2: Analyze the motion of the block
- The distance \( S \) that the block travels relative to the truck is 2 meters.
- The time \( T \) taken for the block to fall off is 2 seconds.
- The initial velocity \( U \) of the block is 0 (since it starts from rest).

Using the equation of motion:
\[ S = Ut + \frac{1}{2} A_2 T^2 \]
where \( A_2 \) is the acceleration of the block relative to the ground.

Substituting the known values:
\[ 2 = 0 \cdot 2 + \frac{1}{2} A_2 (2^2) \]
\[ 2 = 2 A_2 \]
\[ A_2 = 1 \, \text{m/s}^2 \]

### Step 3: Set up the equation of motion for the block
- The block experiences a pseudo force due to the truck's acceleration, which is \( m A_1 \) acting in the backward direction.
- The frictional force \( F \) acts in the forward direction.

Using Newton's second law, we have:
\[ m A_1 - F = m A_2 \]
Substituting \( A_1 = 8 \, \text{m/s}^2 \) and \( A_2 = 1 \, \text{m/s}^2 \):
\[ m \cdot 8 - F = m \cdot 1 \]
\[ F = 8m - 1m = 7m \]

### Step 4: Relate frictional force to normal force
The frictional force \( F \) can also be expressed as:
\[ F = \mu N \]
where \( N \) is the normal force. Since the block is on a horizontal surface, \( N = mg \):
\[ F = \mu mg \]

### Step 5: Equate the two expressions for the frictional force
From the previous steps, we have:
\[ \mu mg = 7m \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \mu g = 7 \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ \mu \cdot 10 = 7 \]
\[ \mu = \frac{7}{10} = 0.7 \]

### Conclusion
The coefficient of sliding friction between the truck and the block is \( \mu = 0.7 \).

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