A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. (Consider `g=10m//s^2`).

Options

A. (a) 4N
B. (b) 16N
C. (c) 20N
D. (d) 22N

Explanation

To solve the problem, we need to find the magnitude of the force applied to the ball while it is in the hand. We will use the work-energy principle to determine this force.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Mass of the ball, \( m = 0.2 \, \text{kg} \)
- Distance moved by the hand while applying the force, \( d = 0.2 \, \text{m} \)
- Additional height gained by the ball after leaving the hand, \( h = 2 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Total Height**:
The total height \( H \) reached by the ball is the sum of the distance moved by the hand and the additional height:
\[ H = d + h = 0.2 \, \text{m} + 2 \, \text{m} = 2.2 \, \text{m} \]

3. **Calculate the Work Done Against Gravity**:
The work done against gravity when the ball rises to height \( h \) is given by:
\[ W_{\text{gravity}} = mgh = 0.2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 2 \, \text{m} = 4 \, \text{J} \]

4. **Calculate the Work Done by the Hand**:
The work done by the hand while applying the force over distance \( d \) is given by:
\[ W_{\text{hand}} = F \cdot d \]
where \( F \) is the force applied by the hand.

5. **Apply the Work-Energy Principle**:
According to the work-energy principle, the total work done on the ball is equal to the change in kinetic energy. Since the ball starts from rest, the initial kinetic energy is 0. The final kinetic energy when it leaves the hand can be expressed as:
\[ W_{\text{hand}} - W_{\text{gravity}} = \Delta KE \]
Since the ball reaches the maximum height, the final kinetic energy at the top is also 0. Thus:
\[ W_{\text{hand}} - 4 \, \text{J} = 0 \]
Therefore:
\[ W_{\text{hand}} = 4 \, \text{J} \]

6. **Substitute and Solve for the Force**:
Now substituting the work done by the hand into the equation:
\[ F \cdot 0.2 \, \text{m} = 4 \, \text{J} \]
Solving for \( F \):
\[ F = \frac{4 \, \text{J}}{0.2 \, \text{m}} = 20 \, \text{N} \]

### Final Answer:
The magnitude of the force applied to the ball is \( \boxed{20 \, \text{N}} \).