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during take - off a 6kg model rocket is burning fuel causing its speed …

Question

during take - off a 6kg model rocket is burning fuel causing its speed to increase at a rate of 6m/s² despite experiencing a 100n drag.
how many forces are acting on the rocket?
what is the magnitude of the net force on the rocket?
what is the strength of the thrust?

Explanation:

Sub - question 1: How many forces are acting on the rocket?
Brief Explanations

To determine the number of forces on the rocket, we analyze the forces acting on it. The rocket is in a vertical motion (take - off). The forces acting on it are: 1. Gravitational force (weight) acting downwards, 2. Drag force (100N) acting opposite to the direction of motion (downwards, as the rocket is moving upwards), 3. Thrust force acting upwards (due to fuel burning which propels the rocket). So, there are 3 forces acting on the rocket.

Step 1: Recall Newton's second law

Newton's second law states that the net force $F_{net}$ acting on an object is equal to the product of its mass $m$ and acceleration $a$, i.e., $F_{net}=m\times a$.

Step 2: Identify the values of mass and acceleration

The mass of the rocket $m = 6\space kg$ and the acceleration $a=6\space m/s^{2}$.

Step 3: Calculate the net force

Substitute the values of $m$ and $a$ into the formula $F_{net}=m\times a$. So, $F_{net}=6\space kg\times6\space m/s^{2}= 36\space N$.

Step 1: Analyze the forces acting on the rocket

Let the thrust be $F_{thrust}$, the gravitational force be $F_{g}=m\times g$ (where $g = 9.8\space m/s^{2}$), and the drag force be $F_{drag}=100\space N$. The net force $F_{net}=F_{thrust}-F_{g}-F_{drag}$. We know from Newton's second law that $F_{net}=m\times a = 36\space N$ (from sub - question 2), $m = 6\space kg$, $g = 9.8\space m/s^{2}$, and $F_{drag}=100\space N$.

Step 2: Calculate the gravitational force

First, calculate the gravitational force $F_{g}=m\times g=6\space kg\times9.8\space m/s^{2}=58.8\space N$.

Step 3: Rearrange the net - force formula to solve for thrust

From $F_{net}=F_{thrust}-F_{g}-F_{drag}$, we can re - arrange it to $F_{thrust}=F_{net}+F_{g}+F_{drag}$.
Substitute the values: $F_{net} = 36\space N$, $F_{g}=58.8\space N$, and $F_{drag}=100\space N$.
$F_{thrust}=36\space N + 58.8\space N+100\space N=194.8\space N\approx195\space N$ (If we take $g = 10\space m/s^{2}$ for approximation, $F_{g}=6\times10 = 60\space N$, then $F_{thrust}=36 + 60+100=196\space N$). But using $g = 9.8\space m/s^{2}$, we have $F_{thrust}=36+58.8 + 100=194.8\space N$.

Answer:

3

Sub - question 2: What is the magnitude of the net force on the rocket?