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### Author Topic: Can energy double?  (Read 14625 times)

#### Tito L. Oracion

• Hero Member
• Posts: 2203
##### Re: Can energy double?
« Reply #45 on: September 28, 2010, 06:11:23 AM »
I think for me the pushing and pulling are two separate energy, then when it hits something it creates another energy effect to that something, therefore i conclude that energy can be trippled!

And! if there is something in that something then there is a corresponding reaction also to that but it can be lesser or higher effect that depends on that something in that something.

#### exnihiloest

• Hero Member
• Posts: 715
##### Re: Can energy double?
« Reply #46 on: September 28, 2010, 10:22:10 AM »
.

#### exnihiloest

• Hero Member
• Posts: 715
##### Re: Can energy double?
« Reply #47 on: September 28, 2010, 10:23:31 AM »
...
The accelerating speed of the iron ball and the car is:
F=ma, F=G-f=1Ã—10-2ï¼8N, a=F/m=8/1=8(m/s2)
...

Wrong calculus. F=ma with a=g applies only in a free fall. Here the ball is prevented by the pulled car from freely falling, then a is much less than g, even if there was no friction.
Without friction, your equation shows that you are expecting the ball to fall with the same acceleration as it was alone, which is obviously false.

Proof:
Let M be the mass of the ball, m the mass of the car, and a their acceleration.
F=(M+m)*a is the force applied to the whole system when its acceleration is a.
F=M*g is the motor force applied to the system (only the ball can move under g, not the car).
It follows that (M+m)*a=M*g then a = g * M/(M+m).
Therefore a<g (even with no losses)

The balance of energy at any time is given by :
M*g*x(t)-1/2*M*vÂ²(t)-1/2*m*vÂ²(t)-Efriction(v) = 0
where v(t) is the speed of both car and ball and x(t) the distance traveled by the ball from its start point, at time t. Efriction is the losses, generally dependent on the car speed or squared speed.