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- anonymous

4. An 800.0 kg roller coaster car is at rest at the top of a 95 m hill. It rolls down the first drop to a height of 31 m. When it travels to the top of the second hill, it is moving at 28 m/s. It then rolls down the second hill until it is at ground level. Draw a picture of the roller coaster drop and calculate the kinetic and potential energy at the top and bottom of each hill.

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- anonymous

Can you help me with this?

- nikato

|dw:1433303891314:dw|
so this would be your diagram. ok?

- anonymous

So how do I do the rest? @nikato

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- nikato

using the formula for potential energy, you can find the PE at point A
remember PE= mgh

- nikato

so 800(10)(95)= 760000 is the potential
and since it's at rest kinetic is 0
this is all at point A

- nikato

find PE at point B
800(10)(31)= 248000
to find kinetic, subtract this from the potential at point A because i'm assuming that energy is conserved
so
760000-248000= 512000 for kinetic at B

- nikato

for C, we are given the velocity so we can find the kinetic energy
\[KE= \frac{ 1 }{ 2 }mv ^{2}\]

- nikato

1/2 (800)(28)^2= 313600
again subtract this from PE at A to find potential at C since energy is conserved
so 760000-313600= 446400

- nikato

assuming that it continues on ground level, all the potential energy at A will be kinetic at D if energy is conserved.
so K at D= 760000
and P=0 since it's at ground level at h=0
remember all my calculations are assuming that you want gravity to be 10 m/s^2

- abb0t

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