A body of mass `10kg` is moving with an initial speed of `20m/s`.The body stops after `5s` due to friction between body and the floor.The value of the coefficient of friction is: (Take acceleration due to gravity `g=10ms^(-2)` )

Options

A. `0.2`
B. `0.4`
C. `0.5`
D. `0.3`

Answer

To solve the problem, we need to find the coefficient of friction (μ) between the body and the floor. Here are the steps to arrive at the solution:

### Step 1: Identify the given values
- Mass of the body (m) = 10 kg
- Initial speed (u) = 20 m/s
- Time taken to stop (t) = 5 s
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the normal force (N)
The normal force (N) acting on the body can be calculated using the formula:
\[ N = m \cdot g \]
Substituting the values:
\[ N = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \]

### Step 3: Determine the deceleration (a)
Since the body comes to a stop, we can use the equation of motion:
\[ v = u + at \]
Where:
- v = final velocity = 0 m/s (since the body stops)
- u = initial velocity = 20 m/s
- t = time = 5 s

Rearranging the equation to find acceleration (a):
\[ 0 = 20 + a \cdot 5 \]
\[ a \cdot 5 = -20 \]
\[ a = -\frac{20}{5} = -4 \, \text{m/s}^2 \]

### Step 4: Relate deceleration to friction
The deceleration (a) is also related to the frictional force (F_friction) acting on the body:
\[ a = \frac{F_{friction}}{m} \]
The frictional force can be expressed as:
\[ F_{friction} = \mu \cdot N \]
Substituting for N:
\[ F_{friction} = \mu \cdot 100 \]

Setting the two expressions for acceleration equal to each other:
\[ -4 = \frac{\mu \cdot 100}{10} \]
\[ -4 = 10\mu \]

### Step 5: Solve for the coefficient of friction (μ)
Now, we can solve for μ:
\[ \mu = -\frac{4}{10} = -0.4 \]

Since the coefficient of friction cannot be negative, we take the absolute value:
\[ \mu = 0.4 \]

### Final Answer
The coefficient of friction (μ) is **0.4**.

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