Q15E - University Physics with Modern Physics

Question

A small 8.00-kg rocket burns fuel that exerts a time-varying upward force on the rocket (assume constant mass) as the rocket moves upward from the launch pad. This force obeys the equation $\mathbf{F}\mathbf{=}\mathbf{A}\mathbf{+}{\mathbf{Bt}}^{\mathbf{2}}$  . Measurements show that at t = 0, the force is 100.0 N, and at the end of the first 2.00 s, it is 150.0 N.

(a) Find the constants A and B, including their SI units. (b) Find the net force on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) 3.00 s after the fuel ignites. (c) Suppose that you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

Step-by-Step Solution

Answer

(a) For the fuel-ignition force obeying the relation ${\text{F = A + Bt}}^{\text{2}}$, $\text{A = 100} \text{N}$and $\text{B = 12.5} \text{kg}·\mathrm{m}/{\mathrm{s}}^4$

(b) (i) Just after fuel ignition the net force on the rocket is 21.6 N and its acceleration is ${\text{2.7m/s}}^2$.(ii) The force on the rocket 3 s after fuel ignition is 134.1 N and its acceleration is $\text{16.76} \mathrm{m}/{\mathrm{s}}^2$.

(c) The acceleration of the rocket 3 s after fuel ignition in outer space is ${\text{26.56m/s}}^2$.

Step 1: Given data

The mass of the rocket is

$\text{m}=8 \text{kg}$

The time-dependent force from fuel ignition is given by the relation

$\text{F}=\text{A}+{\text{Bt}}^{\text{2}}$                    (1)

The force at $\mathrm{t}=0\mathrm{s}$ is

$\text{F}=100 \text{N}$

The force at $\mathrm{t}=2\mathrm{s}$ is

$\text{F}=150 \text{N}$

The acceleration due to gravity is

$\text{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

Step 2: Gravitational force and the second law of motion

The downward gravitational force on a body of mass $m$  is

${\mathbf{F}}_{\mathbf{g}}\mathbf{=}\mathbf{mg}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

The second law of motion relates the net force $F$ , mass $m$  and acceleration $a$ as

$\mathbf{F}\mathbf{=}\mathbf{ma}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(3\right)$

Step 3: The constants A and B

Replace $\mathrm{t}=0\mathrm{s}$ in equation (1) and get

$$\begin{gathered} 100 \text{N}=\text{A}+\text{B}\times {\left(0 \text{s}\right)}^2 \\ \text{A}=100 \text{N} \end{gathered}$$

Replace, $\text{t}=2 \text{s}$ and value of $\mathrm{A}$ in equation (1) and get

$$\begin{gathered} 150 \text{N}=100 \text{N}+\text{B}\times {\left(2 \text{s}\right)}^2 \\ \text{B}=\frac{50 \text{N}}{4 {\text{s}}^{\text{2}}} \\ \text{B}=12.5·\left(1 {\text{s}}^{-2}\right)·\left(1 \text{N}\times \frac{{\displaystyle 1 \text{kg}·\mathrm{m}/{\mathrm{s}}^2}}{1\mathrm{N}}\right) \\ \text{B}=12.5 \text{kg}·\mathrm{m}/{\mathrm{s}}^4 \end{gathered}$$

Thus, A is 100 N and B is $12.5 \text{kg}·\mathrm{m}/{\mathrm{s}}^4$.

Step 4: Net force on the rocket just after fuel ignition

Just after fuel ignition, two forces are acting on the body, the upward force from fuel ignition and the downward force from gravity. The net force is thus

$\begin{array}{rcl}{\mathrm{F}}_{\mathrm{n}}& =& \mathrm{F}\left(\mathrm{t}=0\right)-{\mathrm{F}}_{\mathrm{g}}\\ & =& 100 \text{N}-\mathrm{mg}\\ & =& 100 \text{N}-8 \text{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\\ & =& 21.6 \text{N}\end{array}$

From equation (3), the acceleration of the rocket is

$\begin{array}{rcl}\mathrm{a}& =& \frac{\mathrm{F}}{\mathrm{m}}\\ & =& \frac{21.6 \text{N}}{8 \text{kg}}\\ & =& 2.7·\left(1 {\text{kg}}^{-1}\right)·\left(1 \text{N}\times \frac{ 1\text{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)\\ & =& 2.7\mathrm{m}/{\mathrm{s}}^2\end{array}$

The net force is thus

$\begin{array}{rcl}{\mathrm{F}}_{\mathrm{n}}& =& \text{F}\left(\text{t}=\text{3}\right)-{\text{F}}_{\text{g}}\\ & =& 212.5 \text{N}-\text{mg}\\ & =& 212.5 \text{N}-8 \text{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\\ & =& 134.1 \text{N}\end{array}$

From equation (3), the acceleration of the rocket is

$\begin{array}{rcl}\mathrm{a}& =& \frac{\text{F}}{\text{m}}\\ & =& \frac{134.1 \text{N}}{8 \text{kg}}\\ & =& 16.76·\left(1 {\text{kg}}^{-1}\right)·\left(1 \text{N}\times \frac{1 \text{kg}·{\text{m/s}}^2}{1\mathrm{N}}\right)\\ & =& 16.76 {\text{m/s}}^2\end{array}$

Thus, the net force is 134.1 N and the acceleration is.16.76 m/s².

Step 5: Net force on the rocket 3 s after fuel ignition in space

In space there is no gravity, thus the only force acting on the rocket is the force from fuel ignition. 

The net force after 3 s is thus

$\begin{array}{rcl}{\mathrm{F}}_{\mathrm{n}}& =& \text{F}\left(\text{t}=3\right)\\ & & \text{=}212.5 \text{N}\end{array}$

From equation (3), the acceleration of the rocket is

$\begin{array}{rcl}\mathrm{a}& =& \frac{\text{F}}{\text{m}}\\ & =& \frac{212.5 \text{N}}{8 \text{kg}}\\ & =& 26.56·\left(1 {\text{kg}}^{-1}\right)·\left(1 \text{N}\times \frac{1 \text{kg}·{\text{m/s}}^2}{1\mathrm{N}}\right)\\ & =& 26.56 {\text{m/s}}^2\end{array}$

Thus, the acceleration is 26.56 m/s².