Energy Definition

Oct 2019
9
0
It's sounds suspicious that the difference is due to the limitation of the engines. As I mentioned we can think of 2 cars, one is driving on a road and the sconed is driving on a moving platform, the change in the kinetic energy of the 2 cars will not be the same also they both use the same engine with the same Torque (all this happens in a vacuum so no friction is involved).
 
Last edited:
Oct 2017
676
348
Glasgow
If we will use the full gallon starting from zero velocity, we will reach a velocity of A*1=V.
We can use this gallon to run two spaceships for half an hour, is that case they will reach a speed of A*0.5=0.5V.
Nope.

System 1:
A mass M converts a stored potential energy of \(\displaystyle \Delta E\) to kinetic energy \(\displaystyle KE_1\) over a duration \(\displaystyle \Delta t\) from rest. Calculate the magnitude of constant acceleration and final velocity.

\(\displaystyle KE_1 = \frac{1}{2} M v_1^2 = \Delta E\)

\(\displaystyle v_1 = \sqrt{\frac{2 \Delta E}{M}}\)

\(\displaystyle a_1 = \frac{v_1}{\Delta t} = \frac{1}{\Delta t} \sqrt{\frac{2 \Delta E}{M}}\)

System 2:
A mass M converts a stored potential energy of \(\displaystyle \Delta E / 2\) to kinetic energy \(\displaystyle KE_2\) over a duration \(\displaystyle \Delta t / 2\) from rest. Calculate the magnitude of constant acceleration and final velocity.

\(\displaystyle KE_2 = \frac{1}{2} M v_2^2 = \frac{\Delta E}{2}\)

\(\displaystyle v_2 = \sqrt{\frac{\Delta E}{M}}\)

\(\displaystyle a_2 = \frac{v_2}{\Delta t /2} = \frac{2}{\Delta t} \sqrt{\frac{\Delta E}{M}}\)

Therefore, \(\displaystyle v_1 = \sqrt{2} v_2\) and \(\displaystyle a_1 = \sqrt{2} a_2\)