A roller coaster has a mass of 1200.0kg. The coaster is going 22.0 m/s at the bottom
of the third loop-the-loop that is 2.5m above the ground. Determine the height of
the first hill that is required, assuming the cart is stationary at the top of the first hill
before it falls.

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hamzaahmeds hamzaahmeds · Answer 1 · 2021-07-26T02:32:34+02:00

Answer:

h = 27.17 m

Explanation:

First, we will calculate the total mechanical energy of the system at the bottom point of the third loop:

Mechanical Energy = Kinetic Energy + Potential Energy

[tex]E = \frac{1}{2}mv^2 + mgh[/tex]

where,

E = Total Mechanical Energy = ?

m = mass of the roller coaster = 1200 kg

v = velocity of the roller coaster = 22 m/s

g = acceleration due to gravity = 9.81 m/s²

h = height of roller coaster = 2.5 m

Therefore,

[tex]E = \frac{1}{2}(1200\ kg)(22\ m/s)^2+(1200\ kg)(9.81\ m/s^2)(2.5\ m)\\\\E = 290400 J +29430\ J\\\\E = 319830\ J = 319.83\ KJ[/tex]

Now, the total mechanical energy at the top position of the first hill must also be the same:

[tex]E = \frac{1}{2}mv^2 + mgh[/tex]

where,

v = 0 m/s

h = ?

Therefore,

[tex]319830\ J = \frac{1}{2}(1200\ kg)(0\ m/s)^2+(1200\ kg)(9.81\ m/s^2)(h)\\\\h = \frac{319830\ J}{11772\ N}\\\\[/tex]

h = 27.17 m