Ok, I get it. From 0 m/s to 20 m/s it's 400 J, from 20 m/s to 40 m/s it is 1200 J, it adds up to 1600 J, which is the same as from 0 m/s to 40 m/s. So the 1st formula works. But what about the situation when the initial velocity is 5 m/s, and final is -5 m/s. Would the same formula be used?
It is my understanding that to calculate the change of kinetic energy of an object that speeds up from vi to vf you use this formula:
Change of kinetic energy = 1/2 * m * (vf2 - vi2)
When the initial velocity is 0 m/s I have no problems, but let's say an object that weighs 2 kg speeds up from...
I think i phrased that badly.
What I said there refers to the fact that air resistance increases if the velocity of a falling body icreases too.
And when this force of air resistance reaches a point where it balances the weight force of the body, the acceleration ceases, and the body is from...
Thank you for your responses.
The experiment was conducted in the open air, and the performer lifted the balls as high as his hands could go up (so not very high).
So is it correct to conclude that at this height both balls were accelerating all the way?
And if the height were (much) greater we...
I have seen youtube videos where two balls of the same size and of different masses are dropped from the same height and they hit the ground at the same time.
I understand why this is so: the increase in the mass of a body increases the force of gravity acting on the body, but also decreases...