Question

The first hill of a roller coaster is 42 meters high. The top of the second hill of the one
coaster is 33 meters high. The mass of the roller coaster is 34,500 kg.
a. What is the velocity at the top of the second hill? Ignore friction and assume the
velocity at the top of the first hill is O m/s.
b. What type(s) of energy does the roller coaster have at the top of the second hill?

Answer 1

Answer:

Part a)

Speed of the roller coaster is

$v = 13.3 m/s$

Part b)

Since it is moving with non zero speed at some height above the ground

So we will have

Kinetic energy + Potential energy Both

Explanation:

As we know that there is no friction on the path

So here we can use mechanical energy conservation law

so we will have

Part a)

$U_i  + K_i = U_f + K_f$

$mgh + 0 = \frac{1}{2}mv^2 + mgH$

$34500(9.81)(42) = \frac{1}{2}(34500)v^2 + 34500(9.81)(33)$

$v = 13.3 m/s$

Part b)

Since it is moving with non zero speed at some height above the ground

So we will have

Kinetic energy + Potential energy Both

Answer 2

Answer:

Explanation:

The first hill of a roller coaster is 42 meters high. The top of the second hill of the roller coaster is 33 meters high. The mass of the roller coaster is 34,500 kg