A body of mass 10 kg initially at rest acquires velocity `10 ms^(-1)`. What is the work done ?

Options

A. (a)`-500 J`
B. (b)500 J
C. (c)50 J
D. (d)`-50 J`

Answer

To solve the problem, we need to calculate the work done on a body of mass 10 kg that goes from rest to a velocity of 10 m/s. We can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the mass and initial velocity**:
- Mass (m) = 10 kg
- Initial velocity (u) = 0 m/s (since the body is initially at rest)

2. **Identify the final velocity**:
- Final velocity (v) = 10 m/s

3. **Calculate the initial kinetic energy (KE_initial)**:
- The formula for kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 \]
- Since the initial velocity is 0, the initial kinetic energy is:
\[ KE_{initial} = \frac{1}{2} \times 10 \, \text{kg} \times (0 \, \text{m/s})^2 = 0 \, \text{J} \]

4. **Calculate the final kinetic energy (KE_final)**:
- Now, we calculate the final kinetic energy when the body reaches a velocity of 10 m/s:
\[ KE_{final} = \frac{1}{2} \times 10 \, \text{kg} \times (10 \, \text{m/s})^2 \]
- This simplifies to:
\[ KE_{final} = \frac{1}{2} \times 10 \times 100 = 500 \, \text{J} \]

5. **Calculate the work done (W)**:
- The work done is equal to the change in kinetic energy:
\[ W = KE_{final} - KE_{initial} \]
- Substituting the values we calculated:
\[ W = 500 \, \text{J} - 0 \, \text{J} = 500 \, \text{J} \]

### Final Answer:
The work done on the body is **500 Joules**.

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