Starting from rest a wooden block moves with a velocity of `25ms^(-1)` along a rough ground and comes to rest. Calculate the distance travelled by the wooden block on the rough surface of coefficient of friction 0.25.

Options

A. 50 m
B. 75 m
C. 100 m
D. 125 m

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Initial velocity (u) = 25 m/s
- Final velocity (v) = 0 m/s (since the block comes to rest)
- Coefficient of friction (μ) = 0.25
- Acceleration due to gravity (g) = 9.81 m/s² (standard value)

### Step 2: Calculate the frictional force
The frictional force (F_friction) can be calculated using the formula:
\[ F_{\text{friction}} = \mu \cdot N \]
Where N is the normal force. For a block resting on a horizontal surface, the normal force (N) is equal to the weight of the block (mg).

Thus,
\[ F_{\text{friction}} = \mu \cdot mg \]

### Step 3: Calculate the acceleration (deceleration in this case)
Since the frictional force is the only force acting to stop the block, we can use Newton's second law:
\[ F = ma \]
Where F is the frictional force and a is the acceleration (deceleration in this case). Therefore, we can write:
\[ \mu mg = ma \]

We can cancel m from both sides (assuming m ≠ 0):
\[ \mu g = a \]

Substituting the values:
\[ a = 0.25 \cdot 9.81 \]
\[ a = 2.4525 \, \text{m/s}^2 \]

### Step 4: Use the kinematic equation to find the distance (s)
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Since the block comes to rest, v = 0. Rearranging the equation gives:
\[ 0 = u^2 - 2as \]
\[ 2as = u^2 \]
\[ s = \frac{u^2}{2a} \]

Substituting the known values:
\[ s = \frac{(25)^2}{2 \cdot 2.4525} \]
\[ s = \frac{625}{4.905} \]
\[ s \approx 127.5 \, \text{m} \]

### Final Answer
The distance travelled by the wooden block on the rough surface is approximately **127.5 meters**.

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