Question

At the top of a roller coaster 20 m above the ground the 5 kg car is moving at 2 m/s how fast is it moving at the bottoms of the hill

Answer 1

Answer:

VB = 19.9 [m/s]

Explanation:

In order to solve these problems, we must use the principle of energy conservation. Which tells us that the energy between two points after an instant of time must be equal. This is we have two points A & B, in point A is the highest point while Point B is the lowest

Now we need to identify the types of energy at each point.

For point A

At this point we have two energies, the kinetic energy since the roller coaster moves at a speed of 2 [m/s], in the same way there is potential energy since the roller coaster is 20 [m] above ground level.

For point B

At Point B we only have kinetic energy, since it is located at zero meters with respect to the ground. In this way we can determine the velocity at this point.

And the energy is expressed by means of the following expression:

$E_{A}=E_{B}\\\frac{1}{2} *m*v_{A}^{2} +m*g*h_{A}=\frac{1}{2} *m*v_{B}^{2}$

where:

m = mass = 5 [kg]

VA = velocity of the roller coaster in point A = 2[m/s]

hA = elevation of the roller coaster = 20 [m]

vB = velocity of the roller coaster in point B [m/s]

Now replacing:

$\frac{1}{2}*5*(2)^{2} +(5*9.81*20)=\frac{1}{2}*(5)*v_{B}^{2}\\991=2.5*v_{B}^{2}\\v_{B}x^{2} =991/2.5\\v_{B}=\sqrt{396.4} \\v_{B}=19.9[m/s]$