A force of 10 N inclined to the horizontal at an angle of `60^(@)` acts on a body of mass 2 kg. If the body can move in horizontal direction only find its acceleration.

Options

A. `2.5 m//sec^(2)`
B. `5.5 m//sec^(2)`
C. `4.5 m//sec^(2)`
D. `3.5 m//sec^(2)`

Answer

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

### Step 1: Identify the Given Information
- Force \( F = 10 \, \text{N} \)
- Angle \( \theta = 60^\circ \)
- Mass \( m = 2 \, \text{kg} \)

### Step 2: Resolve the Force into Components
Since the force is inclined at an angle, we need to resolve it into horizontal and vertical components.

- The horizontal component of the force \( F_x \) can be calculated using:
\[ F_x = F \cdot \cos(\theta) \]
- The vertical component of the force \( F_y \) can be calculated using:
\[ F_y = F \cdot \sin(\theta) \]

### Step 3: Calculate the Horizontal Component of the Force
Substituting the values into the equation for \( F_x \):
\[ F_x = 10 \cdot \cos(60^\circ) \]
Using \( \cos(60^\circ) = \frac{1}{2} \):
\[ F_x = 10 \cdot \frac{1}{2} = 5 \, \text{N} \]

### Step 4: Determine the Acceleration
Using Newton's second law of motion, which states that \( F = m \cdot a \), we can rearrange it to find acceleration \( a \):
\[ a = \frac{F_x}{m} \]
Substituting the values:
\[ a = \frac{5 \, \text{N}}{2 \, \text{kg}} = 2.5 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the body is \( 2.5 \, \text{m/s}^2 \).

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