91Ó°ÊÓ

The given data is listed below as:

• The weight of the helicopter is,$\mathrm{w}=2.75×{10}^5\mathrm{N}$
• The time taken by the helicopter is,$t=5.0\mathrm{s}$

The net force is described as the product of the mass and the acceleration of an object. The acceleration of the net force is also described as the derivative of the velocity with respect to the time.

The equation of the position of a training helicopter is expressed as:

$\stackrel{Áåœ}{r}=(0.020\mathrm{m}/{\mathrm{s}}^3)t^3\hat{i}+(2.2\mathrm{m}/\mathrm{s})t\hat{j}-(0.060\mathrm{m}/{\mathrm{s}}^2)t^2\hat{k}$

Here, t is the time taken by the helicopter.

Differentiating the above equation with respect to the time, the velocity of the helicopter can be obtained.

$$\begin{gathered} v=\frac{d\hat{r}\left(t\right)}{d\left(t\right)} \\ =\left(0.06m/s^3\right)t^2\hat{i}+\left(2.2m/s\right)\hat{j}-\left(0.12m/s^2\right)t\hat{k} \end{gathered}$$

Differentiating the above equation with respect to the time, the acceleration of the helicopter can be obtained.

$$\begin{gathered} a=\frac{dv\left(t\right)}{d\left(t\right)} \\ =\left(0.12m/s^3\right)t\hat{i}-\left(0.12m/s^2\right)\hat{k} \end{gathered}$$

The equation of the force of the helicopter can be expressed as:

$F=\frac{w}{g}a$

Here,$w$is the weight of the helicopter, $g$is the acceleration due to gravity and $a$is the acceleration of the helicopter.

Substitute the values in the above equation.

$$\begin{gathered} F=\frac{2.75×{10}^5\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2}\left[\left(0.12\mathrm{m}/{\mathrm{s}}^3\right)\left(5.0\mathrm{s}\right)\hat{\mathrm{i}}-\left(0.12\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{k}}\right] \\ =28061.22\mathrm{N}·{\mathrm{s}}^2/\mathrm{m}\left[\left(0.6\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}-\left(0.12\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{k}}\right] \\ =\left[16836.732\hat{\mathrm{i}}-3367.3464\hat{\mathrm{k}}\right]\mathrm{N} \end{gathered}$$

The equation of the magnitude of the net force can be expressed as:

$F_{net}=\sqrt{{\left(F_x\right)}^2+{\left(F_y\right)}^2+{\left(F_z\right)}^2}$

Here, $F_x$is the force in the $x$direction, $F_y$is the force in the $y$direction and$F_z$ is the force in the $z$direction.

Substitute the values in the above equation.

$$\begin{gathered} F_{net}=\sqrt{{\left(16836.732\right)}^2+{\left(0\right)}^2+{\left(-3367.3464\right)}^2}\mathrm{N} \\ =\sqrt{2.8×{10}^8+1.1×{10}^7}\mathrm{N} \\ =\sqrt{2.9×{10}^8}\mathrm{N} \\ =17029.38\mathrm{N} \end{gathered}$$

Thus, the net force on the helicopter is 17029.38 N.