A block is applied two forces of magnitude 5N each. One force is acting towards East and the other acting along `60^(@)` North of East. The resultant of the two forces (in N) is of magnitude :-

Answer

To find the resultant of the two forces acting on the block, we can follow these steps:

### Step 1: Identify the Forces
We have two forces:
- \( F_1 = 5 \, \text{N} \) acting towards the East.
- \( F_2 = 5 \, \text{N} \) acting at an angle of \( 60^\circ \) North of East.

### Step 2: Draw a Diagram
Draw a coordinate system with East as the positive x-axis and North as the positive y-axis.
- Draw \( F_1 \) along the x-axis (East).
- Draw \( F_2 \) at an angle of \( 60^\circ \) from the x-axis (East) towards the y-axis (North).

### Step 3: Resolve the Forces into Components
To find the resultant force, we need to resolve \( F_2 \) into its x and y components:
- \( F_{2x} = F_2 \cos(60^\circ) = 5 \cos(60^\circ) = 5 \times \frac{1}{2} = 2.5 \, \text{N} \)
- \( F_{2y} = F_2 \sin(60^\circ) = 5 \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \, \text{N} \)

### Step 4: Calculate the Resultant Components
Now, we can find the total components in the x and y directions:
- Total x-component, \( R_x = F_1 + F_{2x} = 5 + 2.5 = 7.5 \, \text{N} \)
- Total y-component, \( R_y = F_{2y} = \frac{5\sqrt{3}}{2} \, \text{N} \)

### Step 5: Calculate the Magnitude of the Resultant Force
Using the Pythagorean theorem, the magnitude of the resultant force \( R \) can be calculated as:
\[ R = \sqrt{R_x^2 + R_y^2} \]
Substituting the values:
\[ R = \sqrt{(7.5)^2 + \left(\frac{5\sqrt{3}}{2}\right)^2} \]
Calculating each term:
\[ R = \sqrt{56.25 + \frac{75}{4}} = \sqrt{56.25 + 18.75} = \sqrt{75} \]

### Step 6: Final Result
Thus, the magnitude of the resultant force is:
\[ R = \sqrt{75} \, \text{N} \]