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

Options

A. ` - 0.5 J `
B. ` - 1.25 J `
C. ` 1.25 J `
D. ` 0.5 J `

Explanation

To solve the problem of finding the work done by gravity on a particle thrown vertically upward, we can follow these steps:

### Step 1: Understand the parameters
- Mass of the particle, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) (since 1 g = 0.001 kg)
- Initial speed, \( u = 5 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Determine the time taken to reach the maximum height
At the maximum height, the final velocity \( v = 0 \). We can use the equation of motion:
\[ v = u + at \]
Substituting the known values:
\[ 0 = 5 - 10t \]
Rearranging gives:
\[ 10t = 5 \quad \Rightarrow \quad t = \frac{5}{10} = 0.5 \, \text{s} \]

### Step 3: Calculate the maximum height reached
We can use the equation for displacement:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ s = 5 \times 0.5 + \frac{1}{2} \times (-10) \times (0.5)^2 \]
Calculating each term:
\[ s = 2.5 + \frac{1}{2} \times (-10) \times 0.25 \]
\[ s = 2.5 - 1.25 = 1.25 \, \text{m} \]

### Step 4: Calculate the work done by gravity
The work done by gravity is given by:
\[ W = F \cdot s \cdot \cos(\theta) \]
Where \( F = mg \) (the weight of the particle), \( s \) is the displacement, and \( \theta \) is the angle between the force and the displacement. Here, \( \theta = 180^\circ \) because the force of gravity acts downward while the displacement is upward.

Calculating the force:
\[ F = mg = 0.1 \, \text{kg} \times 10 \, \text{m/s}^2 = 1 \, \text{N} \]
Now substituting into the work formula:
\[ W = 1 \cdot 1.25 \cdot \cos(180^\circ) = 1.25 \cdot (-1) = -1.25 \, \text{J} \]

### Final Answer
The work done by gravity during the time the particle goes up is \( -1.25 \, \text{J} \).

---