Question

19. (a) what is the average useful power output of a robot that does 5.5 j of useful work in 12.0 hours? (b) working at this rate, how long will it take the robot to lift 3500 kg of bricks 1.8 m to a loading platform?

Answer 1

The average useful power output of robot is 0.0001 W

The time taken by robot to lift bricks is 630000000 sec

Given:

work done by robot = 5.5 j

time taken to do the work = 12.0 hours

weight of bricks = 3500 kg

length of bricks = 1.8 m

To Find:

The average useful power output of robot is

The time taken by robot to lift bricks is

Solution: Useful power output means the electric or mechanical energy made available for use, exclusive of any such energy used in the power production process.

P = W/t

P = 5.5/12*3600

P = 0.0001 W

So, average useful power output of robot is 0.0001 W

t = W/P

W = mgh So t = mgh/P

t = 3500x10x1.8/0.0001

t = 630000000 sec

So, time taken by robot to lift bricks is 630000000 sec

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