Question help a robotic rover on mars finds a spherical rock with a diameter of 1313 centimeters. the rover picks up the rock and lifts it 1515 centimeters straight up. the rock has a specific gravity of 5.255.25. the gravitational acceleration on mars is 3.73.7 meters per second squared. if the robot's lifting arm has an efficiency of 3535% and required 1212 seconds to raise the rock 1515 centimeters, how much power (in watts) did the arm use?
Due to apparent doubling of numbers in the question, assume diameter=13 cm/2=0.065m h=15 cm=0.15m rho=5.25*1000=5250 kg/m^3 eta=35%=0.35 [efficiency] t=12 s g=3.7 m/s^2
mass, m=rho*V=rho*(4pi/3)(r^3) =5250 kg/m^3 * (4pi/3)(0.065)^3 m^3 =1.922 kg
Work done in lifting h=0.15m W=mgh =1.922pi kg * 3.7 m/s^2 * 0.15 m =1.067pi kg (m/s)^2 =1.067pi J
Average power required, with efficiency eta=0.35 P=(W/t)/eta =(1.067pi J )/ (12 s) /0.35 =0.254pi J/s =0.798W (approx.)
