A block of mass 1 kg is placed on a truck which accelerates with acceleration `5ms^(-2)`. The coefficient of static friction between the block and truck is 0.6. The frictional force acting on the block is

Options

A. 5 N
B. 6 N
C. 5.82 N
D. 4.6 N

Answer

To solve the problem, we need to determine the frictional force acting on a block placed on a truck that is accelerating. Let's go through the steps systematically.

### Step 1: Identify the forces acting on the block
The block experiences two main forces:
1. The gravitational force acting downward (weight of the block).
2. The frictional force acting horizontally due to the acceleration of the truck.

### Step 2: Calculate the weight of the block
The weight (W) of the block can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 1 \, \text{kg} \) (mass of the block)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10 \, \text{N} \]

### Step 3: Calculate the maximum static frictional force
The maximum static frictional force (f_s) can be calculated using the formula:
\[ f_s = \mu_s \cdot N \]
where:
- \( \mu_s = 0.6 \) (coefficient of static friction)
- \( N = W = 10 \, \text{N} \) (normal force, which is equal to the weight of the block)

Calculating the maximum static frictional force:
\[ f_s = 0.6 \cdot 10 \, \text{N} = 6 \, \text{N} \]

### Step 4: Calculate the pseudo force acting on the block
Since the truck is accelerating, a pseudo force (F_p) acts on the block in the opposite direction of the truck's acceleration. The pseudo force can be calculated using:
\[ F_p = m \cdot a \]
where:
- \( a = 5 \, \text{m/s}^2 \) (acceleration of the truck)

Calculating the pseudo force:
\[ F_p = 1 \, \text{kg} \cdot 5 \, \text{m/s}^2 = 5 \, \text{N} \]

### Step 5: Determine the frictional force acting on the block
The frictional force acting on the block will be equal to the pseudo force as long as it does not exceed the maximum static frictional force. Since the pseudo force (5 N) is less than the maximum static frictional force (6 N), the frictional force will be equal to the pseudo force.

Thus, the frictional force acting on the block is:
\[ f = F_p = 5 \, \text{N} \]

### Final Answer
The frictional force acting on the block is **5 N**.

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