Simple question on power problems

[Lloydsgurl]

I need a little direction in solving a question on my physics hmw.

We are doing "power problems" and one of the questions asks how long would it take to lift a certain object. The problem gives me the force and mass of the object, plus how high the object will be lifted. But how would I find the amount of time needed?

Any help is appreciated!

 

[nrqed]

I need a little direction in solving a question on my physics hmw.

We are doing "power problems" and one of the questions asks how long would it take to lift a certain object. The problem gives me the force and mass of the object, plus how high the object will be lifted. But how would I find the amount of time needed?

Any help is appreciated!

net force equals m a. With a force F upward, you get F-mg = ma so a=(F/m-g). Once you have the acceleration and the ditsance to cover you can get the time easily (I assume it starts form rest?) using d= 1/2 a t^2.

 

[God64bit]

F=md/s^2
rerange it
s^2 = md/F
Square root it... that what my friend says lol...

 

[nrqed]

F=md/s^2
rerange it
s^2 = md/F
Square root it... that what my friend says lol...

This is *not* what I wrote! First, the acceleration is NOT d/s^2 (if by s you mean the time to get there). And the gravitational force must be taken into account. I gave all the equations in my post.

 

[God64bit]

Friend as in the guy that just climed in my window and answered that question i wasnt tlaking about you.

 

[nrqed]

Friend as in the guy that just climed in my window and answered that question i wasnt tlaking about you.

I understand but I thought that you were implying that your friend had given you the same answer as me. You friend made a few mistakes in his reasoning.

 

[Lloydsgurl]

Well, thanks for the help you two (plus random, window climbing friend, lol).

 

[mrjeffy321]

When lifting an object up against gravity, a certain amount of work must be done. This work can be calculated a couple of different ways.
We know that work equals a force applied over a distance, but work is also a change in energy (in this case, a change in gravitational Potential energy).
W = F * d
W = delta PE

Since you are given the mass of the object and the distance the object is lifted (as well as the force applied), calculating the work needed to lift the object is simply a matter of plugging in the values into the formulas.

In order to calculate power (Energy per unit time) we need to know just how long this work (energy) is being done.

Since you know the force applied to the object as well as the objects mass, you should be able to calculuate the object's acceleration from Newton's second law, [remember to use the net force on the object. Net force = Force applied - weight]
F = ma --> a = F / m
Once you figure out the objects [constant] acceleration from the [constant] force applied to it, you can find the time using a kinematic formula,
d = 1/2 a*t^2
where d is the distance lifted, a is the constant acceleration, and t is time. Solve for t.

Now you have the work done on the object and amount of time the work was done.
Power = Energy / Time

 

[Lloydsgurl]

Thank you mrjeffy321!

This forum (or really, those who contribute such useful information) is about 317 times more helpful than my physics teacher is! I exaggerate not.