On a rough horizontal surface, a body of mass 2 kg is given a velocity of 10 m / s . If the coefficient of friction is 0.2 and `g=10m//s^(2)`, the body will stop after covering a distance of

Options

A. 10 m
B. 25 m
C. 50 m
D. 250 m

Answer

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

### Step 1: Identify the given values
- Mass of the body (m) = 2 kg
- Initial velocity (u) = 10 m/s
- Final velocity (v) = 0 m/s (the body comes to a stop)
- Coefficient of friction (μ) = 0.2
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the normal force (N)
The normal force on a horizontal surface is equal to the weight of the body:
\[ N = m \cdot g \]
Substituting the values:
\[ N = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]

### Step 3: Calculate the frictional force (F_f)
The frictional force can be calculated using the formula:
\[ F_f = \mu \cdot N \]
Substituting the values:
\[ F_f = 0.2 \cdot 20 \, \text{N} = 4 \, \text{N} \]

### Step 4: Calculate the deceleration (a)
Using Newton's second law, the deceleration (a) caused by the frictional force can be calculated as:
\[ F_f = m \cdot a \]
Rearranging gives:
\[ a = \frac{F_f}{m} \]
Substituting the values:
\[ a = \frac{4 \, \text{N}}{2 \, \text{kg}} = 2 \, \text{m/s}^2 \]
Since this is deceleration, we will consider it as negative:
\[ a = -2 \, \text{m/s}^2 \]

### Step 5: Use the kinematic equation to find the distance (s)
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ 0 = (10 \, \text{m/s})^2 + 2(-2 \, \text{m/s}^2)s \]
This simplifies to:
\[ 0 = 100 - 4s \]
Rearranging gives:
\[ 4s = 100 \]
So:
\[ s = \frac{100}{4} = 25 \, \text{m} \]

### Final Answer
The body will stop after covering a distance of **25 meters**.

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