A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters​ [cm]. The rover picks up the rock and lifts it 20 centimeters​ [cm] straight up. The rock has a specific gravity of 4.75. The gravitational acceleration on Mars is 3.7 meters per second squared ​[m divided by s squared​]. If the​ robot's lifting arm has an efficiency of 40​% and required 10 seconds​ [s] to raise the rock 20 centimeters​ [cm], how much power in units of watts​ [W] did the arm​ use?

Respuesta :

Answer:

Power = 0.46 W

Explanation:

Given data:

distance  by which rock lift up is 20 cm

specific gravity of rocks is 4.75

Gravitational acceleration on mars is 3.7 m/sec

Efficiency 40%

we know that specific gravity of rocks

[tex]S.G = \frac{\rho_{rock}}{\rho_{water}}[/tex]

[tex]4.75\times 1000  =\rho_{rock}[/tex]

we know [tex]density = \frac{ mass}{volume}[/tex]

[tex]mass_{rock} = 4.75\times 1000  {\frac{4}{3} \pi (\frac{0.1}{2})^2} = 2.48 kg[/tex]

work done is [tex]W = \frac{mgh}{efficiency}[/tex]

[tex]w = \frac{2.48\times 3.7\times 0.2}{0.4}[/tex]

w = 4.6 j

[tex]so, Power = \frac{w}{t} = \frac{4.6}{10\ sec} = 0.46 W[/tex]