Forces of `5sqrt2` and `6sqrt2N` are acting on a body of mass 1000 kg at an angle of `60^(@)` to each other. Find the acceleration, distance covered and the velocity of the mass after 10 s.

Answer

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

### Step 1: Identify the Forces and the Angle
We have two forces acting on a mass:
- \( F_1 = 5\sqrt{2} \, \text{N} \)
- \( F_2 = 6\sqrt{2} \, \text{N} \)
- The angle \( \theta = 60^\circ \)
- Mass \( m = 1000 \, \text{kg} \)

### Step 2: Calculate the Net Force
To find the net force \( F_{\text{net}} \) when two forces are acting at an angle, we use the formula:
\[ F_{\text{net}} = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos(\theta)} \]

Substituting the values:
\[ F_{\text{net}} = \sqrt{(5\sqrt{2})^2 + (6\sqrt{2})^2 + 2 \cdot (5\sqrt{2}) \cdot (6\sqrt{2}) \cdot \cos(60^\circ)} \]

Calculating each term:
- \( (5\sqrt{2})^2 = 50 \)
- \( (6\sqrt{2})^2 = 72 \)
- \( \cos(60^\circ) = \frac{1}{2} \)
- \( 2 \cdot (5\sqrt{2}) \cdot (6\sqrt{2}) = 60 \)

Putting it all together:
\[ F_{\text{net}} = \sqrt{50 + 72 + 60 \cdot \frac{1}{2}} = \sqrt{50 + 72 + 30} = \sqrt{152} \approx 12.329 \, \text{N} \]

### Step 3: Calculate the Acceleration
Using Newton's second law, the acceleration \( a \) can be calculated as:
\[ a = \frac{F_{\text{net}}}{m} \]
Substituting the values:
\[ a = \frac{12.329}{1000} \approx 0.012329 \, \text{m/s}^2 \]

### Step 4: Calculate the Distance Covered in 10 seconds
Using the formula for distance when initial velocity \( u = 0 \):
\[ S = \frac{1}{2} a t^2 \]
Substituting the values:
\[ S = \frac{1}{2} \cdot 0.012329 \cdot (10)^2 = \frac{1}{2} \cdot 0.012329 \cdot 100 = 0.61645 \, \text{m} \]

### Step 5: Calculate the Velocity after 10 seconds
Using the formula for final velocity:
\[ V = u + at \]
Since \( u = 0 \):
\[ V = 0 + (0.012329 \cdot 10) \approx 0.12329 \, \text{m/s} \]

### Summary of Results
- **Net Force**: \( \approx 12.329 \, \text{N} \)
- **Acceleration**: \( \approx 0.012329 \, \text{m/s}^2 \)
- **Distance Covered in 10 seconds**: \( \approx 0.61645 \, \text{m} \)
- **Velocity after 10 seconds**: \( \approx 0.12329 \, \text{m/s} \)