A bowling machine projects a ball of mass 0.15kg in upward direction. If ball displaced along bowling machine 0.2 m and released. after released from bowling machine ball attain 20 m height then find force exerted by bowling machine on ball

Options

A. 145.5 N
B. 165.5 N
C. 175.5 N
D. 151.5 N

Answer

To solve the problem, we will use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. In this case, we will relate the work done by the bowling machine to the potential energy gained by the ball.

### Step-by-Step Solution:

1. **Identify the given values**:
- Mass of the ball, \( m = 0.15 \, \text{kg} \)
- Displacement along the bowling machine, \( d = 0.2 \, \text{m} \)
- Maximum height attained by the ball after release, \( h = 20 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the total height from the ground**:
The total height \( H \) that the ball reaches from the ground is the sum of the height attained after release and the displacement along the bowling machine:
\[ H = h + d = 20 \, \text{m} + 0.2 \, \text{m} = 20.2 \, \text{m} \]

3. **Calculate the potential energy gained by the ball**:
The potential energy \( PE \) gained by the ball when it reaches the height \( H \) is given by:
\[ PE = m \cdot g \cdot H \]
Substituting the values:
\[ PE = 0.15 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 20.2 \, \text{m} = 30.3 \, \text{J} \]

4. **Relate work done by the bowling machine to potential energy**:
The work done \( W \) by the bowling machine on the ball is equal to the force \( F \) exerted by the machine multiplied by the displacement \( d \):
\[ W = F \cdot d \]
Setting the work done equal to the potential energy gained:
\[ F \cdot 0.2 \, \text{m} = 30.3 \, \text{J} \]

5. **Solve for the force \( F \)**:
Rearranging the equation to find \( F \):
\[ F = \frac{30.3 \, \text{J}}{0.2 \, \text{m}} = 151.5 \, \text{N} \]

### Final Answer:
The force exerted by the bowling machine on the ball is \( 151.5 \, \text{N} \).