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 the block placed on an accelerating truck. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Block
The block of mass \( m = 1 \, \text{kg} \) is on a truck that accelerates with \( a = 5 \, \text{m/s}^2 \). The forces acting on the block include:
- The gravitational force acting downwards, \( F_g = mg \)
- The normal force acting upwards, \( N \)
- The static frictional force acting to the right (in the direction of the truck's acceleration)

### Step 2: Calculate the Gravitational Force
The gravitational force can be calculated using the formula:
\[ F_g = mg \]
where \( g \approx 10 \, \text{m/s}^2 \) (acceleration due to gravity). Therefore,
\[ F_g = 1 \, \text{kg} \times 10 \, \text{m/s}^2 = 10 \, \text{N} \]

### Step 3: Determine the Normal Force
Since the block is not moving vertically, the normal force \( N \) is equal to the gravitational force:
\[ N = F_g = 10 \, \text{N} \]

### Step 4: Calculate the Maximum Static Frictional Force
The maximum static frictional force can be calculated using the coefficient of static friction \( \mu_s \):
\[ F_{\text{friction, max}} = \mu_s \cdot N \]
Given that \( \mu_s = 0.6 \):
\[ F_{\text{friction, max}} = 0.6 \times 10 \, \text{N} = 6 \, \text{N} \]

### Step 5: Calculate the Pseudo Force
When the truck accelerates, a pseudo force acts on the block in the opposite direction of the truck's acceleration. The pseudo force \( F_{\text{pseudo}} \) can be calculated as:
\[ F_{\text{pseudo}} = m \cdot a = 1 \, \text{kg} \times 5 \, \text{m/s}^2 = 5 \, \text{N} \]

### Step 6: Compare the Pseudo Force with Maximum Static Friction
The static frictional force will adjust to counteract the pseudo force as long as it does not exceed the maximum static frictional force. Here, the pseudo force \( 5 \, \text{N} \) is less than the maximum static friction \( 6 \, \text{N} \).

### Step 7: Conclusion
Since the pseudo force \( 5 \, \text{N} \) is less than the maximum static friction \( 6 \, \text{N} \), the frictional force acting on the block will be equal to the pseudo force:
\[ F_{\text{friction}} = 5 \, \text{N} \]

Thus, the frictional force acting on the block is **5 N**.