A block of mass 10 kg starts sliding on a surface with an initial velocity of `9.8 ms^(-1)` . The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is : [use `g=9.8 ms^(-2)` ]

Options

A. 4.9 m
B. 9.8 m
C. 12.5 m
D. 19.6 m

Answer

To solve the problem, we need to determine the distance the block travels before coming to rest due to friction. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Mass of the block, \( m = 10 \, \text{kg} \)
- Initial velocity, \( u = 9.8 \, \text{m/s} \)
- Coefficient of friction, \( \mu = 0.5 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate the frictional force
The frictional force (\( f_k \)) acting on the block can be calculated using the formula:
\[ f_k = \mu \cdot m \cdot g \]
Substituting the values:
\[ f_k = 0.5 \cdot 10 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 49 \, \text{N} \]

### Step 3: Determine the acceleration due to friction
Since the frictional force acts in the opposite direction to the motion, it will cause a deceleration (negative acceleration). We can find the acceleration (\( a \)) using Newton's second law:
\[ f_k = m \cdot a \implies a = \frac{f_k}{m} \]
Substituting the values:
\[ a = \frac{49 \, \text{N}}{10 \, \text{kg}} = 4.9 \, \text{m/s}^2 \]
Since this is a deceleration, we take it as negative:
\[ a = -4.9 \, \text{m/s}^2 \]

### Step 4: Use the kinematic equation to find the distance
We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2a s \]
Where:
- \( v = 0 \, \text{m/s} \) (final velocity when the block comes to rest)
- \( u = 9.8 \, \text{m/s} \) (initial velocity)
- \( a = -4.9 \, \text{m/s}^2 \) (acceleration)

Rearranging the equation to solve for \( s \):
\[ 0 = (9.8)^2 + 2(-4.9)s \]
\[ 0 = 96.04 - 9.8s \]
\[ 9.8s = 96.04 \]
\[ s = \frac{96.04}{9.8} = 9.8 \, \text{m} \]

### Step 5: Conclusion
The distance covered by the block before coming to rest is \( 9.8 \, \text{m} \).

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