A body of mass 1 kg is moving in a vertical circular path of radius 1 m. The difference between the kinetic energies at its highest and lowest position is

Options

A. 20 J
B. 10 J
C. `4sqrt5` J
D. `10(sqrt5-1)` J

Explanation

To solve the problem of finding the difference between the kinetic energies at the highest and lowest positions of a body moving in a vertical circular path, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Mass and Radius**:
- The mass of the body (m) = 1 kg
- The radius of the circular path (r) = 1 m

2. **Understand the Positions**:
- The highest point in the circular path is at a height of 2r from the lowest point.
- The lowest point is at the reference level where potential energy is considered zero.

3. **Calculate the Change in Potential Energy (ΔPE)**:
- The change in height (h) from the lowest point to the highest point is 2r.
- Therefore, h = 2 * 1 m = 2 m.
- The change in potential energy can be calculated using the formula:
\[ \Delta PE = mgh \]
- Substituting the values:
\[ \Delta PE = 1 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2 \, \text{m} = 19.6 \, \text{J} \]

4. **Relate Change in Kinetic Energy to Change in Potential Energy**:
- According to the principle of conservation of mechanical energy, the change in kinetic energy (ΔKE) is equal to the change in potential energy (ΔPE):
\[ \Delta KE = \Delta PE \]
- Therefore, the difference in kinetic energy between the highest and lowest positions is:
\[ \Delta KE = 19.6 \, \text{J} \]

5. **Final Answer**:
- The difference between the kinetic energies at the highest and lowest positions is **19.6 Joules**.