A particle of mass `100 g` is thrown vertically upwards with a speed of `5 m//s`. The work done by the force of gravity during the time the particle goes up is

Options

A. (a)1.25 J
B. (b)0.5 J
C. (c)`-0.5 J`
D. (d)`-1.25 J`

Explanation

To solve the problem, we need to calculate the work done by the force of gravity on a particle of mass `100 g` thrown vertically upwards with an initial speed of `5 m/s`.

### Step-by-Step Solution:

1. **Convert Mass to Kilograms**:
The mass of the particle is given as `100 g`. We need to convert this to kilograms since the standard unit of mass in physics is kilograms (kg).
\[ m = 100 \, \text{g} = \frac{100}{1000} \, \text{kg} = 0.1 \, \text{kg} \]

**Hint**: Remember to convert grams to kilograms by dividing by 1000.

2. **Identify Initial and Final Velocities**:
The particle is thrown upwards with an initial velocity \( u = 5 \, \text{m/s} \). At the highest point of its trajectory, the final velocity \( v = 0 \, \text{m/s} \).

**Hint**: The velocity becomes zero at the highest point of the upward motion.

3. **Calculate Initial Kinetic Energy**:
The initial kinetic energy \( KE_i \) can be calculated using the formula:
\[ KE_i = \frac{1}{2} m u^2 \]
Substituting the values:
\[ KE_i = \frac{1}{2} \times 0.1 \, \text{kg} \times (5 \, \text{m/s})^2 = \frac{1}{2} \times 0.1 \times 25 = 1.25 \, \text{J} \]

**Hint**: Use the kinetic energy formula \( KE = \frac{1}{2} m v^2 \) to find the initial kinetic energy.

4. **Calculate Final Kinetic Energy**:
At the highest point, the final kinetic energy \( KE_f \) is:
\[ KE_f = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.1 \, \text{kg} \times (0 \, \text{m/s})^2 = 0 \, \text{J} \]

**Hint**: The final kinetic energy is zero because the particle momentarily stops at the highest point.

5. **Calculate Work Done by Gravity**:
The work done \( W \) by the force of gravity is equal to the change in kinetic energy:
\[ W = KE_f - KE_i \]
Substituting the values:
\[ W = 0 \, \text{J} - 1.25 \, \text{J} = -1.25 \, \text{J} \]

**Hint**: The work done by gravity is negative because it acts in the opposite direction to the motion of the particle.

6. **Conclusion**:
The work done by the force of gravity during the time the particle goes up is \( -1.25 \, \text{J} \).

**Final Answer**: \( W = -1.25 \, \text{J} \)