Question

A roller coaster has a mass of 1200.0kg. The coaster is going 22.0 m/s at the bottom

Answer 1

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

$E = \frac{1}{2}mv^2 + mgh$

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,

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

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

$E = \frac{1}{2}mv^2 + mgh$

where,

v = 0 m/s

h = ?

Therefore,

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

h = 27.17 m