Chapter 7, Problem 1TP

To determine

The force exerted by the rocket engine on $3.0\text{kg}$ payload.

Answer to Problem 1TP

Option (b), $300\text{N}$.

Explanation of Solution

Given:

Distance traveled with rocket engine firing ( $\text{m}$)	Payload final velocity ( $\text{m}/\text{s}$)
500	310
490	300
1020	450
505	312

Formula used:

The relation between the distance, speed and acceleration is,

  $v^2=u^2+2ad$..... (1)

The relation between force, mass and acceleration is,

  $F=ma$..... (2)

Calculation:

Case 1:

Substituting the given values in equation (1), we get

  $\begin{array}{c}{\left(310\text{m}/\text{s}\right)}^2={\left(0\text{m}/\text{s}\right)}^2+2a_1\left(500\text{m}\right)\\ 2a_1=\frac{{\left(310\text{m}/\text{s}\right)}^2}{\left(500\text{m}\right)}\\ a_1=96.1\text{m}/{\text{s}}^2\end{array}$

Substituting the value in equation (2), we get.

  $\begin{array}{c}{F}_1=\left(3.0\text{kg}\right)\left(96.1\text{m}/{\text{s}}^{\text{2}}\right)\\ =288.3\text{N}\end{array}$

Case 2:

Substituting the given values in equation (1), we get

  $\begin{array}{c}{\left(300\text{m}/\text{s}\right)}^2={\left(0\text{m}/\text{s}\right)}^2+2a_2\left(490\text{m}\right)\\ 2a_2=\frac{{\left(300\text{m}/\text{s}\right)}^2}{\left(490\text{m}\right)}\\ a_2=91.8\text{m}/{\text{s}}^2\end{array}$

Substituting the values in equation (2), we get.

  $\begin{array}{c}{F}_2=\left(3.0\text{kg}\right)\left(91.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =275.4\text{N}\end{array}$

Case 3:

Substituting the given values in equation (1), we get

  $\begin{array}{c}{\left(450\text{m}/\text{s}\right)}^2={\left(0\text{m}/\text{s}\right)}^2+2a_3\left(1020\text{m}\right)\\ 2a_3=\frac{{\left(450\text{m}/\text{s}\right)}^2}{\left(1020\text{m}\right)}\\ a_3=99.26\text{m}/{\text{s}}^2\end{array}$

Substituting the values in equation (2), we get.

  $\begin{array}{c}{F}_3=\left(3.0\text{kg}\right)\left(99.26\text{m}/{\text{s}}^{\text{2}}\right)\\ =297.78\text{N}\end{array}$

Case 4:

Substituting the given values in equation (1), we get

  $\begin{array}{c}{\left(312\text{m}/\text{s}\right)}^2={\left(0\text{m}/\text{s}\right)}^2+2a_4\left(505\text{m}\right)\\ 2a_4=\frac{{\left(312\text{m}/\text{s}\right)}^2}{\left(505\text{m}\right)}\\ a_4=96.38\text{m}/{\text{s}}^2\end{array}$

Substituting the values in equation (2), we get.

  $\begin{array}{c}{F}_4=\left(3.0\text{kg}\right)\left(96.38\text{m}/{\text{s}}^{\text{2}}\right)\\ =289.14\text{N}\end{array}$

The average force exerted on the payload is,

  $\begin{array}{c}F=\frac{288.3\text{N}+275.4\text{N}+297.78\text{N}+289.14\text{N}}{4}\\ =287.655\text{N}\\ \approx 300\text{N}\end{array}$

Conclusion:

Thus, the force exerted by the rocket engine is $300\text{N}$.