A block of mass 2 kg rests on a horizontal surface. If a horizontal force of 5 N is applied on the block, the frictional force on it is `("take", u_(k)=0.4, mu_(s)=0.5)`

Options

A. (a)5N
B. (b)10N
C. (c)8N
D. (d)zero

Answer

To solve the problem step by step, we will analyze the forces acting on the block and determine the frictional force.

### Step 1: Identify the given data
- Mass of the block, \( m = 2 \, \text{kg} \)
- Applied horizontal force, \( F = 5 \, \text{N} \)
- Coefficient of kinetic friction, \( \mu_k = 0.4 \)
- Coefficient of static friction, \( \mu_s = 0.5 \)

### 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 \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).
\[ W = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]

### Step 3: Determine the normal force
Since the block is resting on a horizontal surface, the normal force \( N \) acting on the block is equal to its weight:
\[ N = W = 20 \, \text{N} \]

### Step 4: Calculate the maximum static friction force
The maximum static friction force \( F_s \) can be calculated using the coefficient of static friction:
\[ F_s = \mu_s \cdot N \]
Substituting the known values:
\[ F_s = 0.5 \cdot 20 \, \text{N} = 10 \, \text{N} \]

### Step 5: Compare the applied force with the maximum static friction force
The applied force \( F \) is \( 5 \, \text{N} \), which is less than the maximum static friction force \( F_s = 10 \, \text{N} \). This means that the block will not move, and the frictional force will equal the applied force to prevent motion.

### Step 6: Determine the frictional force
Since the applied force is less than the maximum static friction force, the frictional force \( F_f \) will be equal to the applied force:
\[ F_f = F = 5 \, \text{N} \]

### Final Answer
The frictional force acting on the block is \( 5 \, \text{N} \).

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