# A Physics 11 Elevator Work/Power Problem

• jasmine_01
In summary, the formula for calculating work in an elevator is W = F x d, while the formula for calculating power is P = W/t. The force required to lift an object can be calculated using the formula F = mg. Work is the measure of energy required to move an object, while power is the rate at which work is done. To increase efficiency, one can reduce friction, use a more powerful motor, and optimize weight and balance.
jasmine_01

## 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

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!

## What is the formula for calculating work in an elevator?

The formula for work is W = F x d, where W is the work done, F is the force applied, and d is the distance over which the force is applied.

## What is the formula for calculating power in an elevator?

The formula for power is P = W/t, where P is the power, W is the work done, and t is the time taken to do the work.

## How do you calculate the force required to lift an object in an elevator?

The force required to lift an object in an elevator can be calculated using the formula F = mg, where F is the force, m is the mass of the object, and g is the acceleration due to gravity (9.8 m/s^2).

## What is the difference between work and power in an elevator?

Work is the measure of energy required to move an object over a distance, while power is the rate at which work is done or energy is transferred.

## How can you increase the efficiency of an elevator in terms of work and power?

To increase the efficiency of an elevator, one can reduce the friction in the pulley system, use a more powerful motor, and optimize the weight and balance of the elevator car.

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