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20 March, 01:30

A 1500 kg roller coaster train is at rest at the top of a 65 m drop before it begins to travel down

the hill. At the bottom of the hill how fast is the train moving?

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  1. 20 March, 04:52
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    correct answer: Vmax = 36 m/s

    Explanation:

    Given:

    m = 1,500 kg the mass of the roller coaster

    H = 65 m the height of the hilltop

    Vmax = ? the maximum speed which roller coaster will have at the bottom of the hill

    We will solve this problem using energy conservation laws:

    At the top of the hill, the total roller coaster energy is equal to the maximum potential energy while the kinetic energy is zero:

    Etotal = Epmax and Ek = 0 J

    The formula for calculating potential energy is:

    Ep = m g H we will take g = 10 m/s²

    Epmax = m g H

    At the bottom of the hill, the total roller coaster energy is equal to the maximum kinetic energy while the potential energy is zero:

    Etotal = Ekmax and Ep = 0 J

    The formula for calculating kinetic energy is:

    Ek = m V² / 2

    Ekmax = Etotal = Epmax

    Ekmax = m Vmax² / 2 = m g H

    m Vmax² / 2 = m g H when we divide both sides by mass m we get:

    Vmax² / 2 = g · H ⇒ Vmax² = 2 · g · H

    Vmax² = 2 · 10 · 65 = 1300

    Vmax = √1300 = 36 m/s

    Vmax = 36 m/s

    God is with you!
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