A body moves a distance of 5 m along a straight line under the action of a force 10 N. If the work done is `25sqrt3` joule, the angle which the force makes with the direction of motion of the body is

Options

A. `0^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

To solve the problem step by step, we will use the formula for work done by a force:

### Step 1: Write the formula for work done
The work done (W) by a force (F) when it acts over a displacement (d) at an angle (θ) is given by the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]

### Step 2: Substitute the known values
From the question, we know:
- Work done, \( W = 25\sqrt{3} \) J
- Force, \( F = 10 \) N
- Displacement, \( d = 5 \) m

Substituting these values into the work done formula:
\[ 25\sqrt{3} = 10 \cdot 5 \cdot \cos(\theta) \]

### Step 3: Simplify the equation
Calculating \( 10 \cdot 5 \):
\[ 25\sqrt{3} = 50 \cdot \cos(\theta) \]

### Step 4: Isolate \( \cos(\theta) \)
To find \( \cos(\theta) \), divide both sides by 50:
\[ \cos(\theta) = \frac{25\sqrt{3}}{50} \]

### Step 5: Simplify \( \cos(\theta) \)
This simplifies to:
\[ \cos(\theta) = \frac{\sqrt{3}}{2} \]

### Step 6: Find the angle \( \theta \)
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \]

From trigonometric values, we know:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \theta = 30^\circ \]

### Conclusion
The angle which the force makes with the direction of motion of the body is \( 30^\circ \).

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