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

Answer:

The maximum height could be 10.6 meters.

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 $\frac{mv^2}{2}$ with m the mass, v the velocity and h the height, then:

$\frac{mv_1^2}{2}+mgh_1+W_{f}=\frac{mv_2^2}{2}+mgh_2$

Solving for h_2:

$h_2=\frac{\frac{mv_1^2}{2}+mgh_1+W_{f}-\frac{mv_2^2}{2}}{mg}=\frac{\frac{(1000)(6)^2}{2}+(1000)(9.8)(10)-1700-\frac{(1000)(4.6)^2}{2}}{(1000)(9.81)}$

$h_2=10.6 m$