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. 20 N
B. 22 N
C. 4 N
D. 16N

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass of the ball, \( m = 0.2 \, \text{kg} \)
- Height moved by the hand, \( h_1 = 0.2 \, \text{m} \)
- Additional height the ball goes after leaving the hand, \( h_2 = 2 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the total height
The total height \( h \) that the ball reaches is the sum of the height moved by the hand and the height it travels after leaving the hand:
\[ h = h_1 + h_2 = 0.2 \, \text{m} + 2 \, \text{m} = 2.2 \, \text{m} \]

### Step 3: Calculate the potential energy at the maximum height
The potential energy \( PE \) at the maximum height can be calculated using the formula:
\[ PE = mgh \]
Substituting the values:
\[ PE = 0.2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 2.2 \, \text{m} = 4.4 \, \text{J} \]

### Step 4: Determine the work done by the force
The work done \( W \) by the force while the hand moves the ball is given by:
\[ W = F \cdot s \]
where \( F \) is the force applied and \( s \) is the displacement (height moved by the hand). Here, \( s = 0.2 \, \text{m} \).

### Step 5: Set the work done equal to the potential energy
Since the work done on the ball is converted into potential energy:
\[ F \cdot s = PE \]
Substituting the values we have:
\[ F \cdot 0.2 \, \text{m} = 4.4 \, \text{J} \]

### Step 6: Solve for the force \( F \)
Rearranging the equation to find \( F \):
\[ F = \frac{PE}{s} = \frac{4.4 \, \text{J}}{0.2 \, \text{m}} = 22 \, \text{N} \]

### Conclusion
The magnitude of the force applied by the hand is \( 22 \, \text{N} \).

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