A rocket with a lift-off mass `2xx10^(4)kg` is blasted upwards with an initial acceleration of `5ms^(-2)`. The initial thrust of the blast is `("Take g"=10ms^(-2))`

Options

A. `2xx10^(5)N`
B. `3xx10^(5)N`
C. `4xx10^(5)N`
D. `5xx10^(5)N`

Explanation

To find the initial thrust of the rocket, we can use Newton's second law of motion and the concept of forces acting on the rocket. Here's the step-by-step solution:

### Step 1: Identify the forces acting on the rocket
When the rocket is launched, two main forces act on it:
1. The gravitational force acting downward (weight of the rocket).
2. The thrust force acting upward (force produced by the rocket engines).

### Step 2: Write the equation for the net force
According to Newton's second law, the net force acting on the rocket can be expressed as:
\[ F_{\text{net}} = m \cdot a \]
where:
- \( m \) is the mass of the rocket,
- \( a \) is the acceleration of the rocket.

### Step 3: Calculate the gravitational force
The gravitational force (weight) acting on the rocket can be calculated using the formula:
\[ F_{\text{gravity}} = m \cdot g \]
where:
- \( g \) is the acceleration due to gravity (given as \( 10 \, \text{m/s}^2 \)).

### Step 4: Substitute the values
Given:
- Mass of the rocket, \( m = 2 \times 10^4 \, \text{kg} \)
- Initial acceleration, \( a = 5 \, \text{m/s}^2 \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)

Calculating the gravitational force:
\[ F_{\text{gravity}} = 2 \times 10^4 \, \text{kg} \times 10 \, \text{m/s}^2 = 2 \times 10^5 \, \text{N} \]

### Step 5: Write the equation for thrust
The thrust \( F_{\text{thrust}} \) must overcome both the gravitational force and provide the necessary force for upward acceleration. Thus, we can write:
\[ F_{\text{thrust}} = F_{\text{gravity}} + F_{\text{net}} \]
Substituting \( F_{\text{net}} = m \cdot a \):
\[ F_{\text{thrust}} = F_{\text{gravity}} + m \cdot a \]

### Step 6: Calculate the thrust
Substituting the values:
\[ F_{\text{net}} = 2 \times 10^4 \, \text{kg} \times 5 \, \text{m/s}^2 = 1 \times 10^5 \, \text{N} \]

Now substituting into the thrust equation:
\[ F_{\text{thrust}} = 2 \times 10^5 \, \text{N} + 1 \times 10^5 \, \text{N} = 3 \times 10^5 \, \text{N} \]

### Final Answer
The initial thrust of the blast is:
\[ F_{\text{thrust}} = 3 \times 10^5 \, \text{N} \]

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