Question

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?

Answer 1

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

$S.G = \frac{\rho_{rock}}{\rho_{water}}$

$4.75\times 1000 =\rho_{rock}$

we know $density = \frac{ mass}{volume}$

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

work done is $W = \frac{mgh}{efficiency}$

$w = \frac{2.48\times 3.7\times 0.2}{0.4}$

w = 4.6 j

$so, Power = \frac{w}{t} = \frac{4.6}{10\ sec} = 0.46 W$