Answer to Question #91506 in Mechanics | Relativity for Baren

Question

For the rollercoaster x=22 m. A loaded 519 kg roller coaster cart has 8.8104*10^4 Kinetic energy at point x. And 1.8627*10^5 of KE at point Y. What is the total mechanical energy at point x and Y? What is the height of the track at point Y?

Answer 1

1. The total mechanical energy E of the object is equal to the sum of the kinetic energy K and the gravitational potential energy U

$$E=K+U$$

where kinetic energy is

$$K=\frac{m{{v}^{2}}}{2}$$

m is the mass and v is the speed of the object, gravitational potential energy is

$$U=mgh$$

g=9.8 m/s²  is the acceleration of gravity, h is the height of the object.

We are given the kinetic energy at point x: ${{K}_{x}}=8.8104\cdot {{10}^{4}}\,J$

Find the gravitational potential energy at point x (h=x=22 m):

$${{U}_{x}}=mgx=519\text{ }kg\cdot 9.8\,m/{{s}^{2}}\,\cdot 22\,m \\ =\text{111},896\,J=11.1896\cdot {{10}^{4}}J$$

Find the total mechanical energy at point X: ${{E}_{x}}={{K}_{x}}+{{U}_{x}}$. Substituting the known values, we get

$${{E}_{x}}=8.8104\cdot {{10}^{4}}\,J+11.1896\cdot {{10}^{4}}J\\=20\cdot {{10}^{4}}J=2\cdot {{10}^{5}}J$$

Since there is no energy loss, then mechanical energy is conserved and the total mechanical energy at point Y is equal to the total energy at point X. Thus

$${{E}_{x}}={{E}_{y}}=2\cdot {{10}^{5}}J$$

2. Find the height of the track at point Y.

The total mechanical energy of the object at point Y is ${{E}_{y}}={{K}_{y}}+{{U}_{y}}$ , then ${{U}_{y}}={{E}_{y}}-{{K}_{y}}$ . Substituting the known values, we get

$${{U}_{y}}=2\cdot {{10}^{5}}J-1.8627\cdot {{10}^{5}}J=\text{0}\text{,1373}\cdot {{10}^{5}}J\\=\text{1}\text{.373}\cdot {{10}^{4}}J$$

Recall that ${{U}_{y}}=mgy$, then height y of the track at point Y is

$$y=\frac{{{U}_{y}}}{mg}$$

Substituting the known values, we get

$$y=\frac{\text{1}\text{.373}\cdot {{10}^{4}}J}{519\text{ }kg\cdot 9.8\,m/{{s}^{2}}}=2.7\,m$$