Q:

if we have two acrobats one is 3 m above the ground and has weight of 1000 N,second is standing on a lever on ground and has a weight of 300 N.
If the first acrobat jump,is the second acrobat reach a 7m height ?

- hamad

Kuwait

- hamad

Kuwait

A:

Hamad -

Well, I don't want to do your homework for you, but I can help to show you how this problem works... I think the easiest way to approach this is in terms of energy, namely potential energy. The thing about energy is that it's always conserved. That is, even though it may get transfered from one object to another, the total amount is always the same.

In this problem, you've really got two different spots in time to think about. The first one has the first acrobat in the air, just about to start falling. But he's not falling*yet*, so all he has is
potential energy. The second point is when the second acrobat is in the
air as high as he's going to get - he's halfway in between moving
upwards and moving downwards, so he's not actually moving at all. This
means that all he's got is potential energy.

The equation for (gravitational) potential energy is mass*gravity*height. So the energy at first is equal to the first acrobat's mass, times the acceleration of gravity (9.81m/s^2), times how high up he is. The energy at the end is equal to the second acrobat's mass, times gravity, times how high up*he* is. Your
problem makes it a little bit easier, since it gives you the acrobats'
weights instead of masses. The weight equals the mass times the
acceleration of gravity, so you get two equations for the potential
energy at two different times, and you know that both have to be equal:

Energy at first = Acrobat 1's Weight * Acrobat 1's Height

Energy at last = Acrobat 2's Weight * Acrobat 2's Height

Now you can just set these two equations equal to each other and solve to see how high up the second acrobat can go.

-Tamara

Well, I don't want to do your homework for you, but I can help to show you how this problem works... I think the easiest way to approach this is in terms of energy, namely potential energy. The thing about energy is that it's always conserved. That is, even though it may get transfered from one object to another, the total amount is always the same.

In this problem, you've really got two different spots in time to think about. The first one has the first acrobat in the air, just about to start falling. But he's not falling

The equation for (gravitational) potential energy is mass*gravity*height. So the energy at first is equal to the first acrobat's mass, times the acceleration of gravity (9.81m/s^2), times how high up he is. The energy at the end is equal to the second acrobat's mass, times gravity, times how high up

Energy at first = Acrobat 1's Weight * Acrobat 1's Height

Energy at last = Acrobat 2's Weight * Acrobat 2's Height

Now you can just set these two equations equal to each other and solve to see how high up the second acrobat can go.

-Tamara

*(published on 10/22/2007)*