A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is `5m//s^2`, the frictional force acting on the block is…………newtons.

Options

A. 10N
B. 5N
C. 2.5N
D. 20N

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass of the block (m) = 1 kg
- Coefficient of static friction (μ) = 0.6
- Acceleration of the truck (a) = 5 m/s²
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the normal force (N)
The normal force acting on the block is equal to its weight since it is on a horizontal surface.
\[ N = m \cdot g \]
\[ N = 1 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \]

### Step 3: Calculate the maximum static frictional force (F_s)
The maximum static frictional force can be calculated using the formula:
\[ F_s = \mu \cdot N \]
\[ F_s = 0.6 \cdot 9.8 \, \text{N} = 5.88 \, \text{N} \]

### Step 4: Calculate the applied force (F_a)
The applied force on the block due to the acceleration of the truck can be calculated using Newton's second law:
\[ F_a = m \cdot a \]
\[ F_a = 1 \, \text{kg} \cdot 5 \, \text{m/s}^2 = 5 \, \text{N} \]

### Step 5: Compare the applied force with the maximum static frictional force
Now we compare the applied force (F_a) with the maximum static frictional force (F_s):
- \( F_a = 5 \, \text{N} \)
- \( F_s = 5.88 \, \text{N} \)

Since the applied force (5 N) is less than the maximum static frictional force (5.88 N), the block will not slip, and the frictional force acting on the block will be equal to the applied force.

### Step 6: Conclusion
Thus, the frictional force acting on the block is:
\[ F_{\text{friction}} = F_a = 5 \, \text{N} \]

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

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