Question

A spaceship lifts off vertically from the Moon, where$\mathbf{g}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. If the ship has an upward acceleration of$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$as it lifts off, what is the magnitude of the force exerted by the ship on its pilot, who weighs 735 Non Earth?

Short answer

Force exerted by ship on the pilot is 195 N.

Step-by-step solution

Given

1. Gravitational acceleration on the moon, $g_m=1.6\mathrm{m}/{\mathrm{s}}^2$,
2. The upward acceleration is $a=1.0\mathrm{m}/{\mathrm{s}}^2$and
3. The weight of the pilot on the earth is $W_{onearth}=735\mathrm{N}$.

Understanding the concept Newton’s law

acceleration.

Use Newton’s second law to find the force. Using given values, we can find the mass of the pilot. The net force on the pilot is the sum of applied force and the force of gravity.

Formulae:

$$\begin{gathered} \mathbf{F}\mathbf{=}\mathbf{ma} \\ \mathbf{W}\mathbf{=}\mathbf{mg} \end{gathered}$$

Here,F is force,m is mass,a is acceleration,W is weight,g is gravitational acceleration.

Calculate the magnitude of the force exerted by the ship on its pilot

The calculation for the force exerted by the ship on the pilot:

The weight of the person on earth is,

$$\begin{gathered} W=m{g}_e \\ =735\mathrm{N} \end{gathered}$$

$$\begin{gathered} m=\frac{W}{g_e} \\ =\frac{735\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =75\mathrm{kg} \end{gathered}$$

Net force is,

$$\begin{gathered} F_{net}=F_{app}-W_{moon} \\ ma=F_{app}-\left(m{g}_m\right) \\ \left(75\mathrm{kg}\mathrm{x}1.0\mathrm{m}/{\mathrm{s}}^2\right)=F_{app}\left(75\mathrm{kg}\mathrm{x}1.6\mathrm{m}/{\mathrm{s}}^2\right) \\ {F}_{app}=\left(75\mathrm{kg}\mathrm{x}1.0\mathrm{m}/{\mathrm{s}}^2\right)+\left(75\mathrm{kg}\mathrm{x}1.6\mathrm{m}/{\mathrm{s}}^2\right) \\ {F}_{app}=195\mathrm{N} \end{gathered}$$

Therefore, the force exerted by ship on the pilot is 195 N.