Question

Two electric motors drive two elevators of equal mass in a three-story building 10 meters tall. Each elevator has a mass of 1,000 kg. The first elevator can do this work in 5.0 seconds. Calculate the power output of the first motor.
2.0 x 102 W
2.0 x 103 W
2.0 x 104 W
2.0 x 105 W

Answer 1

Answer:

2.0 x 104 W

Explanation:

Same process as the person above but use significant figures for the problem.

Answer 2

Answer:

2.0 * 10^4 W

Explanation:

We will be using these two equations:

$W=m*g*h$

Power = ΔW/Δt

PART 1----------------------

First, we find the work done by the motor.

If you are in the OW physics course, you will already know the value of W from the previous question.

If you don't know W yet, use the equation

$W=m*g*h$

where

W = work;

m = mass of elevator;

g = acceleration due to gravity;

h = height.

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We are given the values for m, g, and h:

m = 1000kg;

g = acceleration due to gravity = 9.8m/s^2;

h = 10m.

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We plug the known values into the equation and solve for W:

$W=m*g*h$

$W = 1000kg * 9.8m/s^2 * 10m$

$W = 9.8 * 10^{4} J$

PART 2----------------------

Now we use the formula

Power = ΔW/Δt

where

ΔW = change in work

Δt = change in time

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We are given a value of t, which is the interval over which the work takes place:

Δt = 5.0 seconds

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Now we can plug in our values for W and t and solve:

$Power = \frac{9.8 * 10^{4} J}{5.0 sec}$

$Power = 19600W$

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In scientific notation:

1.96 * 10^4 W

Accounting for the maximum of two significant digits:

2.0 * 10^4 W

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