A person has a box of weight 20 kg. The energy of the box, when the person runs with a constant velocity of 2`ms^(-1)` along with the box behind the bus, is given as

Options

A. 20J
B. 40 J
C. 80J
D. zero

Answer

To solve the question, we need to find the energy of the box when a person runs with it at a constant velocity of 2 m/s. The energy we are looking for is primarily the kinetic energy since the potential energy is negligible in this scenario.

### Step-by-Step Solution:

1. **Identify the mass of the box**:
The mass (m) of the box is given as 20 kg.

2. **Identify the velocity of the box**:
The velocity (v) at which the person is running with the box is given as 2 m/s.

3. **Calculate the Kinetic Energy (KE)**:
The formula for kinetic energy is:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 20 \, \text{kg} \times (2 \, \text{m/s})^2 \]

4. **Calculate \(v^2\)**:
\[ (2 \, \text{m/s})^2 = 4 \, \text{m}^2/\text{s}^2 \]

5. **Substitute \(v^2\) into the KE formula**:
\[ KE = \frac{1}{2} \times 20 \, \text{kg} \times 4 \, \text{m}^2/\text{s}^2 \]

6. **Calculate the Kinetic Energy**:
\[ KE = \frac{1}{2} \times 80 \, \text{kg m}^2/\text{s}^2 = 40 \, \text{Joules} \]

7. **Determine the Potential Energy (PE)**:
Since the box is assumed to be at ground level (or close to it), the height (h) is negligible, and thus:
\[ PE = mgh = 20 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0 \, \text{m} = 0 \, \text{Joules} \]

8. **Calculate the Total Energy**:
The total energy (E) is the sum of kinetic energy and potential energy:
\[ E = KE + PE = 40 \, \text{Joules} + 0 \, \text{Joules} = 40 \, \text{Joules} \]

### Final Answer:
The energy of the box when the person runs with a constant velocity of 2 m/s is **40 Joules**.