A block of mass 5 kg is kept on a horizontal floor having coefficient of friction 0.09. Two mutually perpendicular horizontal forces of 3 N and 4 N this block. The accelation of the block is ( take, `g=10 ms^(-2)`)

Options

A. zero
B. `0.1 ms^(-2)`
C. `0.2 ms^(-2)`
D. `0.3 ms^(-2)`

Explanation

To solve the problem step by step, we will follow these calculations:

**Step 1: Identify the forces acting on the block.**
- The block has two perpendicular forces acting on it: \( F_x = 3 \, \text{N} \) and \( F_y = 4 \, \text{N} \).

**Step 2: Calculate the resultant force.**
- Since the forces are perpendicular, we can use the Pythagorean theorem to find the resultant force \( F' \):
\[ F' = \sqrt{F_x^2 + F_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{N} \]

**Step 3: Calculate the normal force (N).**
- The normal force can be calculated using the formula:
\[ N = m \cdot g \]
where \( m = 5 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \):
\[ N = 5 \cdot 10 = 50 \, \text{N} \]

**Step 4: Calculate the maximum force of friction (F_max).**
- The maximum force of friction can be calculated using the coefficient of friction (\( \mu \)):
\[ F_{\text{max}} = \mu \cdot N \]
Given \( \mu = 0.09 \):
\[ F_{\text{max}} = 0.09 \cdot 50 = 4.5 \, \text{N} \]

**Step 5: Calculate the net force (F).**
- The net force acting on the block is the resultant force minus the maximum frictional force:
\[ F = F' - F_{\text{max}} = 5 - 4.5 = 0.5 \, \text{N} \]

**Step 6: Calculate the acceleration (a) of the block.**
- Using Newton's second law, \( F = m \cdot a \):
\[ a = \frac{F}{m} = \frac{0.5}{5} = 0.1 \, \text{m/s}^2 \]

Thus, the acceleration of the block is \( 0.1 \, \text{m/s}^2 \).