A body of mass 10kg is placed on a smooth inclined plane making an angle of `30^(@)` with the horizontal, the component of the force of gravity trying to move the body down the inclined plane is `[g=9.8m//s^(2)]`

Options

A. 98 N
B. 49 N
C. 10 N
D. 5 N

Answer

To solve the problem of finding the component of the gravitational force acting on a body of mass 10 kg placed on a smooth inclined plane at an angle of 30 degrees, we can follow these steps:

### Step 1: Identify the weight of the body
The weight (W) of the body can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 10 \, \text{kg} \) (mass of the body)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 10 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

### Step 2: Break down the weight into components
The weight acts vertically downwards. We need to find the component of this weight that acts down the incline. This can be done using trigonometric functions. The component of the weight acting down the incline (F) is given by:
\[ F = W \cdot \sin(\theta) \]
where \( \theta = 30^\circ \).

### Step 3: Calculate the component of the force
Substituting the values into the equation:
\[ F = 98 \, \text{N} \cdot \sin(30^\circ) \]

We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]

Thus:
\[ F = 98 \, \text{N} \cdot \frac{1}{2} = 49 \, \text{N} \]

### Conclusion
The component of the force of gravity trying to move the body down the inclined plane is:
\[ \boxed{49 \, \text{N}} \]

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