Question about the Conservation of Energy

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SUMMARY

The discussion centers on the application of the law of conservation of energy to calculate the velocity of a cart rolling down a ramp and flying off a table. The user correctly identifies the formula mgh1 = mgh2 + 1/2mv² and simplifies it to v = √(2g(h1 - h2)). However, a critical error arises in the height calculations, where the user mistakenly considers h1 as 5 meters instead of the correct total height of 8 meters (5m ramp + 3m table). This miscalculation leads to incorrect results in determining the cart's velocity.

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nicholz
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If a cart rolled down a ramp 5 meters high and flew off a a table at 3 meters high, what would its velocity be?

please let me know where i go wrong, i was using the law of conservation of energy

mgh1=mgh2+1/2mv2
mg(h1-h2)=1/2mv2
v=square root of 2g(h1-h2)

where h1=5 and h2=3

i plugged in numbers for the variables and i can't seem to get it to work out. is the equation i came up with correct?
edit: i was thinking that it should come out to the square root of 2gh. would that be right though? can i consider the release height to be the "0" height here?
 
Last edited:
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Yes your equation is correct, though since the ramp is 5m high and then the table is 3m high, shouldn't h1=8m?
 

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