How to find work when you have force and time

alicia113

I have an at home lab. That tells me to run up a flight of stairs. My flight of stairs is 3 m (vertical height). To walk up the stairs it takes me 6.98 seconds. To run up the stairs it takes me 2.88 seconds. I need to find the work that is being done. I weigh 118 pounds that is 526 N. How do I come about this I know work is

W=(F)(D)

So do I need to put my time in seconds into something. But what? Please help me!!

CFede

As you mention [itex]W=\vec{F}\cdot \vec{d}[/itex], time doesn't appear there, and that's because the work done by a force is independent of the time length of the force action.

The meassurement of time it takes you to go upstairs is not really that useful. Try to approach the problem in a different way.

Hint: When you go upstairs the work you do becomes energy. Think about it, How much more energy do you have after you have gone upstairs? where does this energy comes from?

CWatters

What CFede said. Look at it from the point of view of energy gained.

There is an interesting conclusion.

alicia113

Is it

F x D
--------
t
??????

CFede

Hi again alicia, i think you are very confused. The time here, is good for nothing. It doesn't matter how long it takes you to go upstairs, whether you can do t in 3 minutes or 3 hours is irrelevant, the work done is the same. You must approach the problem in a different manner, let me try to explain:

When the energy of a body changes, that energy change must ahve come from somewhere. If at some moment, a body A has more energy that at an earlier time, that "extra" energy it has, he must have obtained from something else, this means, that some work had to be done to the body A in order to grant him that "extra" energy.

When you go upstairs, your energy increases, more specifically, your gravitational potencial energy increases. This increase in the potential energy occurs because of the work done, which means, the increase in the potential energy is equivalent to the work done, this is:

[itex]W=\Delta E=E_f-E_i=mgh_f-mgh_i=mg\Delta h[/itex]

In this way, you can relate the work done, with the energy difference. As you can see time plays no role in this whole thing. You must think of the problem in terms of energy changes.

Another comment too. This equation you mention [itex]W=\vec{F}\cdot\vec{d}[/itex] is true for constant (or average) forces only. Since this is not the case, you shouldn't use that.

Hope you understand a little better now. If not, ask again.

alicia113

Ok so I under stood the question wrong. It says I need to find my power rating. Is that correct? For my previous Anwser?

I was thinking power rating was work ... It's not work correct ?

Mark M

You can think of power as being the rate of energy change, whereas work is the total change in energy, independent of time. So, if we let [itex] \Delta W [/itex] be the work performed, the equation for the average power is [tex] P_{avg} = \frac { \Delta W} { \Delta t} [/tex] So, calculate the work using the formula provided by CFede, and then divide by time.

alicia113

Ok thanks!!

So it will be
W=FxD
= 526N x 3m
=1578J

Pavrg= 1578J/6.98m
= 226.98. ( what's the unit)

alicia113

6.98s I'm sorry!

Mark M

Well, you should use CFede's formula to calculate the work (e.g. The change in gravitational potential energy.).

Mark M

Also, the unit of power is Joules per second (J/s).

CWatters

Not quite.. It's...

Pavrg= 1578J/6.98S
= 226.98 J/S (eg Joules per second)

1 J/S = 1 Watt

so

Pavrg = 226.98 Watts

If you prefer horses then 750W=1Hp so 226.98 is about 0.3 Hp

alicia113

So that's my power than?

alicia113

Thank you so much!!

So my power for 2.88s is 0.7 hp or 529.53 watts

CWatters

1578j/2.88s = 548w