Tpr sw physics passage 7

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crazy person

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Can someone please explain to me #1 and #2

I'm confused why on #1 the answer says that PE is on top of the trajectory when the passage says that at the top of the trajectory, the KE is 600J. shouldn't we use the KE?
Even though I know that once the KE is released, it transforms to PE but according to the passage it states that at the top of the trajectory it has KE 600 J. so shouldn't we use the 600J ?

ON #2: again, why is the KE used on the horizontal direction? shouldn't PE be used instead for the horizontal direction?

I'm so confused.

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Can someone please explain to me #1 and #2

I'm confused why on #1 the answer says that PE is on top of the trajectory when the passage says that at the top of the trajectory, the KE is 600J. shouldn't we use the KE?
Even though I know that once the KE is released, it transforms to PE but according to the passage it states that at the top of the trajectory it has KE 600 J. so shouldn't we use the 600J ?

ON #2: again, why is the KE used on the horizontal direction? shouldn't PE be used instead for the horizontal direction?

I'm so confused.
I can't see the problem, as I don't have your book, but let's see what I can clarify from what you've given so far:
KE is dependent on the object's overall velocity, which has both x and y components. Therefore, you can separate it out into "KE from horizontal mvmt" and "KE from vertical mvmt". You MUST do this in order to properly answer these questions

Horizontal:
Because horizontal velocity is constant, the amount of KE of the horizontal mvmt is also constant. (KE = ½mv² where v is horizontal velocity).
Vertical:
Because vertical velocity changes over time, it is NOT constant. We know that the change in potential gravitational energy is dependent on height y Ug = mgy
and that at the top of the arc, vertical velocity = 0, so KE from vertical mvmt is also 0, and Ug will therefore be at its highest value.

So, for Q1, if the object is at the top of its arc, we know that its vertical velocity is 0, so the KE from vertical movement is also 0. Therefore, if the object has any KE, it is ONLY from the horizontal velocity. As horizontal KE is constant, we know that this is the same amount of horizontal KE as it had at the beginning...subtract that 600J from the initial total KE and you'll have the amount of initial vertical KE, which will have all been converted to Ug by the top of the flight...giving you your Ug value.

For Q2, there is no Potential energy for horizontal motion. Gravitational potential energy is only affected by vertical movement...if you move a ball up and stop supporting it, it will fall back down. If you move a ball to the left and stop touching it, it will not fall back to the right, because there is no energy 'stored' in the ball to move it that direction.
 
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