Can energy be decomposed?

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WuliDancer
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TL;DR
Force is a vector, so there is synthesis and decomposition of force; velocity is a, so there is synthesis and decomposition of velocity. Energy can be produced by doing work (F·S), and it can also be expressed as velocity (Ek=mV²). So can energy be decomposed?
Let's construct a model: a small ball with mass m is thrown on a horizontal plane with a V, and the direction of the velocity makes an angle θ with the horizontal plane. What is the maximum height it can reach? (ignoring air resistance) We that the kinetic energy of an object will not be completely converted into gravitational potential energy. The velocity in the horizontal direction will not be affected by any acceleration (the of acceleration is external force, and the object is not subjected to force in the horizontal direction when it is flying in the air). Therefore, it will not be converted gravitational potential energy. The amount of velocity in the vertical direction that can be converted into height depends on the velocity at the highest point - obviously 0, so the is completely converted into height.

Let's try:
Initial kinetic energy: E1=½mV²=½mV²sin²θ+½mV²cos²θ
You will find that sin²θ+cos²θ=1 seems to be a coincidence. Let's continue
Final kinetic energy: E2=½mV²cos²θ
Change in kinetic energy: δE=-½mV²sin²θ
The change in kinetic energy is exactly the opposite of change in gravitational potential energy, so
Change in gravitational potential energy: δE=½mV²sin²θ
δE=mgδh
Solution for δh=½V²sin²θ/g
Do you agree with this solution?
 
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WuliDancer said:
So can energy be decomposed?
Since it doesn't have components, it can not. Energy is a scalar quantity, not a vector: it has a magnitude but no direction.

WuliDancer said:
You will find that sin²θ+cos²θ=1 seems to be a coincidence
It is not a coincidence. You could consider it a consequence, though (divide Pythagoras ##a^2+ b^2 = c^2## by ##c^2##) :wink:

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I think what you did is easier understood by writing the velocity as a vector, ##\vec{v}##. The kinetic energy is then ##\frac 12m\vec{v}\cdot\vec{v}## which you can expand as ##\frac 12m(v_x^2+v_y^2+v_z^2)##.

So you can't break energy down into components in the sense of a vector, but you can write an energy associated with each component of a vector and they will add to the total energy (as long as you work in an orthogonal basis, anyway). Since it's only ##v_z## that's changed by gravity, it's only the ##\frac 12mv_z^2## that changes.
 
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