A block of mass 1 kg is resting on a rough inclined plane which rises·3 in every 5. What is the minimum force required to move the block up the inclined plane ifµ == 0.5? (g`=10m//s^2`)

Options

A. 10 N
B. 8 N
C. 6 N
D. 4 N

Explanation

To solve the problem of finding the minimum force required to move a block of mass 1 kg up a rough inclined plane, we can follow these steps:

### Step 1: Determine the dimensions of the incline
The incline rises 3 units for every 5 units of horizontal distance. We can visualize this as a right triangle where:
- The height (opposite side) = 3
- The base (adjacent side) = 5

Using the Pythagorean theorem, we can find the length of the incline (hypotenuse):
\[ \text{Length of incline} = \sqrt{(3^2 + 5^2)} = \sqrt{(9 + 25)} = \sqrt{34} \]

### Step 2: Calculate the angle of inclination (θ)
We can find the angle θ using the sine function:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{\sqrt{34}} \]
And using the cosine function:
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{\sqrt{34}} \]

### Step 3: Calculate the weight of the block
The weight (W) of the block is given by:
\[ W = mg \]
Where:
- \( m = 1 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)

Thus,
\[ W = 1 \times 10 = 10 \, \text{N} \]

### Step 4: Resolve the weight into components
The weight of the block can be resolved into two components:
1. **Parallel to the incline**:
\[ W_{\parallel} = W \sin \theta = 10 \cdot \frac{3}{\sqrt{34}} \]
2. **Perpendicular to the incline**:
\[ W_{\perpendicular} = W \cos \theta = 10 \cdot \frac{5}{\sqrt{34}} \]

### Step 5: Calculate the normal force (N)
The normal force (N) acting on the block is equal to the perpendicular component of the weight:
\[ N = W_{\perpendicular} = 10 \cdot \frac{5}{\sqrt{34}} \]

### Step 6: Calculate the frictional force (f)
The frictional force (f) opposing the motion is given by:
\[ f = \mu N \]
Where \( \mu = 0.5 \):
\[ f = 0.5 \cdot \left(10 \cdot \frac{5}{\sqrt{34}}\right) = \frac{25}{\sqrt{34}} \]

### Step 7: Calculate the minimum force (F)
The minimum force required to move the block up the incline must overcome both the gravitational component down the incline and the frictional force:
\[ F = W_{\parallel} + f = 10 \cdot \frac{3}{\sqrt{34}} + \frac{25}{\sqrt{34}} \]
Combining the terms:
\[ F = \frac{30 + 25}{\sqrt{34}} = \frac{55}{\sqrt{34}} \]

### Step 8: Calculate the numerical value of F
To find the numerical value of \( F \):
\[ F \approx \frac{55}{5.831} \approx 9.43 \, \text{N} \]

### Conclusion
The minimum force required to move the block up the inclined plane is approximately **9.43 N**.

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