anonymous
  • 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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Can you help me with this?
nikato
  • nikato
|dw:1433303891314:dw| so this would be your diagram. ok?
anonymous
  • anonymous
So how do I do the rest? @nikato

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nikato
  • nikato
using the formula for potential energy, you can find the PE at point A remember PE= mgh
nikato
  • 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
  • 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
  • nikato
for C, we are given the velocity so we can find the kinetic energy \[KE= \frac{ 1 }{ 2 }mv ^{2}\]
nikato
  • 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
  • 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
  • abb0t
@Jhannybean

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