Work Done yadda yadda These problems are killing me

In summary, the traveler did work for 6.10 seconds on the escalator, due to the elevator's movement. The escalator did work for 3030 Joules.
  • #1
Antepolleo
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Ok, I'm having trouble with yet another work-related problem. Here it is:

A traveler at an airport takes an escalator up one floor. The moving staircase would itself carry him upward with vertical velocity component 0.8 m/s between entry and exit points separated by height 10.0 meters. However, while the escalator is moving, the hurried traveler climbs the steps of the escalator at a rate of 2.8 steps per second. Assume that the height of each step is 0.3 meters.

a) Determine the amount of work done by the traveler during his escalator ride, given that his mass is 80.0 kg

b) Determine the work the escalator motor does on this person.

For a, I summed the man's and the escalator's velocities and used the W = Fd equation. I'm not sure how to go about b.

I was wrong for a, so I'm really up a creek on this one. Any help is greatly appreciated.
 
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  • #2
For a, I summed the man's and the escalator's velocities and used the W = Fd equation.
You did? Even though W= Fd has nothing to do with the velocities?
It doesn't matter how FAST you apply a force, only that you apply it.
I would have been inclined to note that the total work done is the
change in potential energy. You are given that the man's mass is 80 kg. and that he has moved up a vertical distance of 10 m: change in potential energy= work done= (80)(9.81)(10)= 7848 Joules.

For B, look at the work the man has done. HERE'S where the speeds come in. The elevator is moving up with a vertical speed of 0.8 m/s and the man is walking up at a vertical speed of 2.8 STEPS per second while each step is 0.3 m so that is 0.84 m/s, for a total vertical velocity of 1.64 m/s. The point of that is that it will have taken him 10/1.64= 6.10 seconds to go up the elevator.

Now you can finish the problem in either of two ways:

Directly: Since the elevator is moving upward at 0.8 m/s, it will have carried him up 0.8(6.10)= 4.88 m (he walked up the other 5.12 m)
and thus did work of (80)(9.81)(4.88)= 3830 Joules on the man.

Indirectly: The man was walking upwards at 0.84 m/s so in the 6.1 seconjds, he walked up (.84)(6.1)= 5.12 m (we saw that before) and thus did (80)(9.81)(5.12)= 4018 J of work himself. The elevator did the other 7848- 4018= 3030 J of work.
 
  • #3
Thank you a thousand times for your help. Really I don't know what I was thinking, using that equation. I should learn to consider if a solution makes sense before I apply it.

Thanks again!
 

1. What is "work done" in scientific terms?

"Work done" is a measure of the energy transferred when a force is applied to an object and causes it to move a certain distance. It is calculated by multiplying the force applied by the distance the object moves in the direction of the force.

2. How is work done related to force and distance?

Work done is directly proportional to the force applied and the distance the object moves. This means that as the force or distance increases, so does the work done.

3. Can you give an example of calculating work done?

Sure! Let's say you push a box with a force of 10 Newtons for a distance of 5 meters. The work done would be 50 Joules (10 N x 5 m = 50 J).

4. Is work done the same as power?

No, work done and power are not the same. Work done is a measure of energy transferred, while power is a measure of how quickly work is done. Power is calculated by dividing work done by the time it takes to do the work.

5. Why is understanding work done important in science?

Understanding work done is important because it allows us to quantify and analyze the effects of forces on objects. It is also a fundamental concept in many scientific fields, such as physics and engineering, and is essential in understanding concepts such as energy and motion.

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