Question

A 1000kg roller coaster begins at 10m tall hill with initial velocity of 6m/s and travels down until a second hill. 1700J is transformed to thermal engird by friction. In order for rollercoaster to get up 2nd hill it must have velocity of 4.6m/s or less at the top. What is the maximum height the second hill could be

Answer 1

The maximum height could be 10.6 meters.

Step-by-step explanation:

For this kind of exercise, we use the general principle for conservation of mechanical energy (E) that states:

`E_1+W_f=E_2` (1)

That means the mechanical energy an object has on a point 2 should be equal to the mechanical energy on a point 1 plus the energy transformed into heat due friction denoted as Wf (It is negative because is lost). In our case point 1 is the point where the roller coaster begins and point 2 is at the second hill. Tola mechanical energy is the sum of potential gravitational energy and kinetic energy, so (1) is :

`K_(1)+U_(1)+W_(f)=K_(2)+U_(2)`

with K the kinetic energy and U the potential energy, remember potential energy is mgh and kinetic energy is
`(mv^2)/(2)` with m the mass, v the velocity and h the height, then:

`(mv_1^2)/(2)+mgh_1+W_(f)=(mv_2^2)/(2)+mgh_2`

Solving for h_2:

`h_2=((mv_1^2)/(2)+mgh_1+W_(f)-(mv_2^2)/(2))/(mg)=(((1000)(6)^2)/(2)+(1000)(9.8)(10)-1700-((1000)(4.6)^2)/(2))/((1000)(9.81))`

`h_2=10.6 m`