A cylinder of 10 kg is sliding in a plane with an initial velocity of 10 m / s . If the coefficient of friction between the surface and cylinder is 0.5 then before stopping, it will cover. `(g=10m//s^(2))`

Options

A. 10m
B. 7.5m
C. 5m
D. 2.5m

Explanation

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

### Step 1: Identify the given data
- Mass of the cylinder (m) = 10 kg
- Initial velocity (u) = 10 m/s
- Coefficient of friction (μ) = 0.5
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the frictional force
The frictional force (F_f) acting on the cylinder can be calculated using the formula:
\[ F_f = \mu \cdot m \cdot g \]

Substituting the values:
\[ F_f = 0.5 \cdot 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 \]
\[ F_f = 0.5 \cdot 100 \]
\[ F_f = 50 \, \text{N} \]

### Step 3: Calculate the acceleration due to friction
The frictional force causes a deceleration (negative acceleration) on the cylinder. Using Newton's second law:
\[ F = m \cdot a \]
We can rearrange this to find acceleration (a):
\[ a = \frac{F_f}{m} \]

Substituting the values:
\[ a = \frac{50 \, \text{N}}{10 \, \text{kg}} \]
\[ a = 5 \, \text{m/s}^2 \]

Since this is a deceleration, we will consider it as negative:
\[ a = -5 \, \text{m/s}^2 \]

### Step 4: Use the kinematic equation to find the distance covered before stopping
We will use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- v = final velocity = 0 m/s (when the cylinder stops)
- u = initial velocity = 10 m/s
- a = acceleration = -5 m/s²
- s = distance covered

Rearranging the equation to solve for s:
\[ 0 = (10)^2 + 2(-5)s \]
\[ 0 = 100 - 10s \]
\[ 10s = 100 \]
\[ s = \frac{100}{10} \]
\[ s = 10 \, \text{m} \]

### Final Answer
The cylinder will cover a distance of **10 meters** before stopping.

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