You are an engineer designing a roller coaster. You want the cars to be moving a

Question

You are an engineer designing a roller coaster. You want  the cars to be moving at 5.00m/s at the top of the big hill which is 35.0m above the ground. The bottom of  the "dip" just before the big hill is 6.25m above the ground. Measurements show that 12.5% of the initial kinetic energy of the cars will be converted to heat and sound between the "dip" and the top of the big hill. How fast should the cars be moving at the botom of the dip? Given: 1.00m/s=2.24MPH

            

             Vf=5.00m/s

Ignore the 45m and 30m.

the 45m---------> is 35.00m

the 30m---------> is 6.25m

Given

6.25m

35.00m

Vf=5.00m/s

Vo=?

You are an engineer designing a roller coaster. You want the cars to be moving at 5.00m/s at the top of the big hill which is 35.0m above the ground. The bottom of the "dip" just before the big hill is 6.25m above the ground. Measurements show that 12.5% of the initial kinetic energy of the cars will be converted to heat and sound between the "dip" and the top of the big hill. How fast should the cars be moving at the botom of the dip? Given: 1.00m/s=2.24MPH

Explanation / Answer

Hey! I made a mistake. Here's the correct answer.

Kinetic Energy at dip = 1/2 m Vo^2

Potential energy at dip = m g * 6.25

Kinetic energy at top = 1/2 m * 5^2 = 1/2 m * 25

Potential energy at top = m g * 35

Kinetic Energy lost from dip to top = 0.125 * (1/2 m Vo^2)

Initial Energy = m g * 6.25 + (1/2 m Vo^2)

Final Energy = m g * 35 + ( 1/2 m * 25 )

Initial Energy - Final Energy = Loss = 0.125 (1/2 m Vo^2)

1/2 m ( Vo^2 - 25 ) - mg * 28.75 = 1/2 m (0.125 Vo^2)

1/2 (0.875 Vo^2) = 12.5 + g * 28.75 = 12.5 + 9.8 * 28.75 = 294.25

Vo^2 = 672.57

Vo = 25.934 m/s

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