A body falls freely under gravity. Its speed is v when it has lost an amount U of the gravitational energy. Then its mass is

Options

A. `(Ug)/v^(2)`
B. `U^(2)/g`
C. `(2U)/v^(2)`
D. `2 U gv^(2)`

Answer

To solve the problem, we need to relate the gravitational potential energy lost by the body to its kinetic energy gained as it falls freely under gravity.

### Step-by-Step Solution:

1. **Understanding the Energy Conservation**:
When a body falls freely under the influence of gravity, its total mechanical energy (the sum of potential energy and kinetic energy) remains constant. As the body falls, it loses gravitational potential energy and gains kinetic energy.

2. **Initial Energy**:
Let's denote the initial height from which the body falls as \( h \). The initial potential energy (PE) of the body at height \( h \) is given by:
\[ PE_{\text{initial}} = mgh \]
where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity.

3. **Energy Lost**:
According to the problem, the body loses an amount \( U \) of gravitational potential energy as it falls. Therefore, the potential energy at a height \( h' \) (after falling) can be expressed as:
\[ PE_{\text{final}} = PE_{\text{initial}} - U = mgh - U \]

4. **Kinetic Energy Gained**:
The kinetic energy (KE) gained by the body when it has fallen and reached speed \( v \) is given by:
\[ KE = \frac{1}{2} mv^2 \]

5. **Applying Conservation of Energy**:
Since the total mechanical energy is conserved, the loss in potential energy equals the gain in kinetic energy:
\[ U = KE \]
Thus, we have:
\[ U = \frac{1}{2} mv^2 \]

6. **Solving for Mass \( m \)**:
Rearranging the equation to solve for \( m \):
\[ m = \frac{2U}{v^2} \]

### Final Answer:
The mass of the body is given by:
\[ m = \frac{2U}{v^2} \]