Recent content by aris1

In an horizontal surface the force of the weight is perpendicular to the displacement so no work against gravity is done if that's what you mean 'gravity doesn't play a role'. Only work against friction is done.

Yes...after the power is removed the two cars have exactly the same KE and negative acceleration acting on them. The cars will travel exactly the same distance and will stop after exactly the same time.

You already know the answer. Positive net force(force applied greater than the weight) causes acceleration and negative net force(force applied less than the weight) causes deceleration.

Hey...I didn't go that far.I assumed that the energy is also linear and positive only thrust(deceleration only due to gravity). Anyway,thank you very much for your time and thinking. So in the hovering case where no energy goes to the rocket(or no work is produced) the whole energy goes into...

That third case is what bothers me the most. In my opinion since the staring speed(v1) and the ending speed(v2) is zero then the average acceleration(a) is zero too(a=(v2-v1)/t=(0-0)/t=0) and the motion is practically identical with constant speed motion.The energy expenditure has to be the same.

I believe when we talk about the force and the energy of the engine we mean exactly the the force and the energy of the exhaust. Even if we use the conservation of the momentum in all the three cases where the average acceleration is zero(the rocket(or helicopter) stands still in the...

Hey... my question was a lot more simple and my example maybe was a little misleading. I just would like to know if it's OK to assume that in cases where the average acceleration is zero the energy expenditure is always the same for the same duration. The cases where the average acceleration...

Thanks.That's exactly what I thought. Do you also believe that equal amounts of energy are required when the rocket starts and ends with zero speed(accelerates and then decelerates,average acceleration=zero)?

I'm discussing about the energy needed for the engine. In my opinion when the average acceleration is zero(like in the cases where the rocket/helicopter starts and ends at rest OR just hovering in the air) the average force is equal with the weight and the energy expenditure for the same...

Hey...it's just a fictional example.I guess the helicopter example is better. So you agree that the energy expenditure is the same in any case with the only difference the efficiency of their engines?

Thanks. So you also agree that in the B case where the rocket starts and ends at rest(a case close to reality) which means that the average acceleration is zero the rocket would burn the same fuels as if it was simply hovering? So you're saying that when the motion is with constant speed(or...

Hi all, I want to ask something with the fear of sound silly to most of you. A)Let's say we have two identical rockets and the air resistance is excluded. The first rocket travels upwards with 100mph while the second travels upwards with 50mph.It's obvious that after an hour the faster...

So you want to find the peak value of force if,for example,let the weight fall free and then at the last inches you try to stop it immediately? Maybe you can calculate the speed at this point either by the equation v=gt (if you count the time) or by the equation v=(2gh)^.5 and for h you can put...

Are we talking about weight lifting here?If so I don't think we can consider that we accelerate for all the way up or down. In fact the weight always starts and ends at rest.That means the average acceleration is always zero and the force is always equal with the weight(F=mg) both in the way up...