A block of mass 20kg is pushed with a horizontal force of 90N. If the coefficient of static & kinetic friction are 0.4 & 0.3, the frictional force acting on the block is: (g`=10ms^(-2)`)

Options

A. 90 N
B. 80 N
C. 60 N
D. 30 N

Answer

To solve the problem, we need to determine the frictional force acting on a block of mass 20 kg when a horizontal force of 90 N is applied. We are given the coefficients of static and kinetic friction as 0.4 and 0.3, respectively.

### Step-by-Step Solution:

**Step 1: Calculate the weight of the block.**
The weight (W) of the block can be calculated using the formula:
\[ W = m \times g \]
Where:
- \( m = 20 \, \text{kg} \) (mass of the block)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 20 \, \text{kg} \times 10 \, \text{m/s}^2 = 200 \, \text{N} \]

**Step 2: Determine the normal force (N).**
Since the block is on a horizontal surface and there are no vertical forces acting on it apart from its weight and the normal force, the normal force (N) is equal to the weight of the block:
\[ N = W = 200 \, \text{N} \]

**Step 3: Calculate the maximum static friction force (f_s).**
The maximum static friction force can be calculated using the formula:
\[ f_s = \mu_s \times N \]
Where:
- \( \mu_s = 0.4 \) (coefficient of static friction)

Calculating the maximum static friction:
\[ f_s = 0.4 \times 200 \, \text{N} = 80 \, \text{N} \]

**Step 4: Compare the applied force with the maximum static friction.**
The applied force is 90 N, which is greater than the maximum static friction force of 80 N. This means the block will start moving, and we will now consider kinetic friction.

**Step 5: Calculate the kinetic friction force (f_k).**
The kinetic friction force can be calculated using the formula:
\[ f_k = \mu_k \times N \]
Where:
- \( \mu_k = 0.3 \) (coefficient of kinetic friction)

Calculating the kinetic friction:
\[ f_k = 0.3 \times 200 \, \text{N} = 60 \, \text{N} \]

**Step 6: Conclusion.**
The frictional force acting on the block, once it starts moving, is the kinetic friction force:
\[ f_k = 60 \, \text{N} \]

### Final Answer:
The frictional force acting on the block is **60 N**.