A block of 10kg is pulled at a constant speed on a rough horizontal plane by a force of `20N`. Calculate the coefficient of friction ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass of the block (m) = 10 kg
- Force applied (F) = 20 N
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the normal force (N)
The normal force (N) acting on the block on a horizontal plane is equal to the weight of the block, which can be calculated using the formula:
\[ N = m \cdot g \]
Substituting the values:
\[ N = 10 \, \text{kg} \cdot 9.8 \, \text{m/s}² = 98 \, \text{N} \]

### Step 3: Understand the relationship between frictional force and applied force
Since the block is moving at a constant speed, the net force acting on it is zero. This means that the applied force (F) is equal to the frictional force (f):
\[ F = f \]

### Step 4: Write the formula for the coefficient of friction (μ)
The coefficient of friction (μ) is defined as the ratio of the frictional force (f) to the normal force (N):
\[ \mu = \frac{f}{N} \]

### Step 5: Substitute the known values into the equation
Since we know that \( f = F \):
\[ \mu = \frac{F}{N} \]
Substituting the values we have:
\[ \mu = \frac{20 \, \text{N}}{98 \, \text{N}} \]

### Step 6: Calculate the coefficient of friction
Now, performing the division:
\[ \mu = \frac{20}{98} \approx 0.204 \]

### Step 7: Round the result
Rounding to two decimal places, we get:
\[ \mu \approx 0.20 \]

### Final Answer:
The coefficient of friction is approximately **0.20**.

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