Question

A 2550 pound roller coaster starts from rest and is launched such that it crests a 119 ft high hill with a speed of 57 mph. The roller coaster travels 583 ft in reaching the top of the hill and there is a constant drag force of 100 pounds. Determine launch energy.

Answer 1

To solve this problem we will apply the concepts related to energy conservation. For this purpose we will have that all the changes occurred in the energy change will be equivalent to the change in the potential and kinematic energies of the body. At the same time we will consider that the change in the final energy of the system will be reflected in the work of the system, therefore,

`\Delta E = KE+PE`

`E_i - Fd = (1)/(2)mv^2+Wh`

Here,

F = Force

m = mass

v = Velocity

h = Height

d = Distance

`W = mg \rightarrow \text{Weight/Force Weight}`

`E_i =(1)/(2)mv^2+Wh+Fd`

`E_i = (1)/(2) ((W)/(g))v^2 +Wh+fd`

Replacing we have,

`E_i = (1)/(2) ((2550pounds)/(32.2ft/s^2))(57mi/h((1.467ft/s)/(1mi/h)))+(2550pounds)(119ft)+(100pounds)(583ft)`

`E_i =365061ft\cdot lb`

Therefore the launch energy is 365061ft-lb