A Physics 11 Elevator Work/Power Problem

AI Thread Summary
The discussion revolves around calculating the work and power involved in an elevator transporting a 70 kg person and a 500 kg elevator to a height of 400 m in 4 minutes. The potential energy gained by the system is calculated using the formula Ep = mgh, resulting in 2,234,400 J, which translates to a power output of 9,310 W when divided by the time taken. The efficiency of the motor is noted to be 25%, leading to a required input power of 37,240 W to account for energy losses. The conversation emphasizes the importance of incorporating both the elevator's and the person's mass in the calculations and understanding the impact of efficiency on motor performance. Overall, the calculations are confirmed to be accurate and well-structured.
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Homework Statement



An elevator takes a 70 kg person from street level to the top of a 400 m sky-scraper in 4 minutes. The mass of the elevator is 500 kg.

a] At what rate is the elevator plus the person gaining potential energy?
b] What power is the elevator plus the person experiencing as it climbs?
c] The motor driving the lift has an efficiency of only 25%. At what rate is the motor working as the lift is going up?

Homework Equations



How can the elevator calculation be incorporated with the person?
How can the efficiency affect the calculation for the work of the motor? [/B]

The Attempt at a Solution



mgh = (70kg)(9.8m/s^2)(400m)
Work = Force*Displacement
rate of Work = Power

Efficiency = Workout/Workin*100%
0.25 * (400m*500kg*9.8m/s^2)/240s
Power = Work/Time
4 minutes = 240 s
 
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jasmine_01 said:

The Attempt at a Solution



mgh = (70kg)(9.8m/s^2)(400m)
Work = Force*Displacement
rate of Work = Power

Efficiency = Workout/Workin*100%
0.25 * (400m*500kg*9.8m/s^2)/240s
Power = Work/Time
4 minutes = 240 s

Thanks for using the homework template - unfortunately that cannot be taken for granted. However, the equations and formulas you show in the last point are not really the attempt of a solution, although you need them to solve the problem (its more for point 2).

jasmine_01 said:

Homework Equations



How can the elevator calculation be incorporated with the person?
How can the efficiency affect the calculation for the work of the motor?

Regarding your questions in point 2:

1) If the motor lifts the elevator - does it care if a person is in the elevator or if it just weighs 570 instead of 500 kg?
2) The motor cannot transform the entire power received from the power plant into potential energy. A certain part of the electric power is transformed into heat (by friction, ...) and is counted as loss. With an efficiency of 25 %, according to your equation, how large are the losses?

With respect to these hints, please try to correct your equations and show us your attempt to solve the problem step by step (that means the relevant equations shall be brought in the correct order to solve the questions 1-3 one after the other).
 
My attempt at finding the solutions:
(1) As part a) question asks about potential energy, the equation most relevant in this case would be Ep = mgh - where m is the mass, g is the gravitational force, and h is the height. It doesn't matter if the motor lifts the elevator itself or along with the person so for the calculation, my first step would be finding the potential enegy. With the potential energy I can find the rate of change of potential energy when divided by the amount of time it takes.

Step 1:
mass = 500 + 70 = 570 kg
gravity = 9.8 m/s^2
height = 400 m

Step 2: Multiplication (570)(9.8)(400) = 2,234,400 J

Step 3: 2,234,400/240 = 9310 W
(2) As part b) question asks about power, the equation Power = Work/Time would be most relevant in this case. As it involves work, the equation Work = Force×Displacement would also be relevant as well. As time is known and displacement is known, I would need to find everything else that is unknown to solve for Power. (I am assuming the force of gravity/weight of elevator and the person is the force)
Step 1: Using the equation Weight=mg where m is the mass of the object (s) and g is the force of gravity

Step 2: Force=(500+70=570kg)*(9.8m/s^2)=5,586N

Step 3: Using the equation Work=Force*Displacement, I can find Work=5,586N*400m=2,234,400 J

Step 4: As Power=Work/Time, Power=2,234,400 J/240s=9310 W
(3)As part c) of the question asks about efficiency, I would need to incorporate what I know about efficiency. The equation most revelant to this question would be Workout/Workin*100% = Efficiency. I assume that 25% could be made equal to Workout/Workin*100%.
Step 1: Using 25% given in the question with the equation (25%=Workout/Workin*100%)

Step 2: Cancel out the 100 by dividing 100 on both sides ---> it becomes 0.25 = Workout/Workin

Step 3: 9310 J = Workout ---> it becomes 0.25=9310/Workin

Step 4: Solve for Workin ---> it becomes 9,310/0.25=37,240 W
 
Perfect, technically and formally. Well done!
 
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