A block of mass 10 kg is moving horizontally with a speed of 1.5 `ms^(-1)` on a smooth plane. If a constant vertical force 10N acts on it, the displacement of the block from the point of application of the force at the end of 4 second is

Options

A. 5m
B. 20 m
C. 12 m
D. 10 m

Explanation

To solve the problem, we need to determine the displacement of the block after 4 seconds, considering both its horizontal and vertical movements.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the block, \( m = 10 \, \text{kg} \)
- Initial horizontal speed, \( u_x = 1.5 \, \text{m/s} \)
- Vertical force acting on the block, \( F = 10 \, \text{N} \)
- Time, \( t = 4 \, \text{s} \)

2. **Calculate the Horizontal Displacement:**
- The horizontal displacement can be calculated using the formula:
\[ \text{Displacement} = \text{Speed} \times \text{Time} \]
- Substituting the values:
\[ a = u_x \times t = 1.5 \, \text{m/s} \times 4 \, \text{s} = 6 \, \text{m} \]

3. **Calculate the Vertical Acceleration:**
- Using Newton's second law, the acceleration \( a_y \) due to the vertical force can be calculated as:
\[ F = m \times a_y \implies a_y = \frac{F}{m} = \frac{10 \, \text{N}}{10 \, \text{kg}} = 1 \, \text{m/s}^2 \]

4. **Calculate the Vertical Displacement:**
- The vertical displacement can be calculated using the formula:
\[ b = u_y \times t + \frac{1}{2} a_y t^2 \]
- Since the initial vertical velocity \( u_y = 0 \):
\[ b = 0 \times 4 + \frac{1}{2} \times 1 \, \text{m/s}^2 \times (4 \, \text{s})^2 = \frac{1}{2} \times 1 \times 16 = 8 \, \text{m} \]

5. **Calculate the Resultant Displacement:**
- The resultant displacement \( d \) can be calculated using the Pythagorean theorem:
\[ d = \sqrt{a^2 + b^2} = \sqrt{(6 \, \text{m})^2 + (8 \, \text{m})^2} \]
- Calculating:
\[ d = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m} \]

### Final Answer:
The displacement of the block from the point of application of the force at the end of 4 seconds is **10 meters**.