Question

A roller coaster starts from rest at the top of an 18m hill. The car travels to the bottom of the hill and continues up the next hill that is 10.0m high. How fast is the car moving at the top of the 10.0m hill if friction is ignored?

Answer 1

Answer:

12.53 m/s

Explanation:

Neglect friction and air resistance and take the 10 m hill as the point of reference for potential energy, by the law of energy conservation, when the car travels from 18m hill to 10m hill,  its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv^2/2$

where m is the car mass and h = 18 - 10 = 8 m is the net vertical distance traveled, v is the car velocity at the 10m hill, which we are looking for

we can divide by m on both sides

$gh = v^2/2$

Let g = 9.81 m/s2. We can solve for v

$v^2 = 2gh = 2*9.81*8 = 156.96$

$v = \sqrt{156.96} = 12.53 m/s$

Answer 2

The change of velocity with respect to time is called acceleration. The acceleration depends on the velocity and the speed.

According to the question, the data is as follows:-

If we neglect friction and air resistance and take the 10 m hill as the point of reference for potential energy.

The law of energy conservation, when the car travels from 18m hill to 10m hill,  its potential energy is converted to kinetic energy.

Hence, $E_p=E_k$

$mgh = \frac{1}{2}mv^2$

The vertical distance is 8m.

by solving the equation, we will get

$gh= \frac{1}{2}v^2$

After putting the value and solving it, we will get:-

$v = \sqrt{2gh}\\\\=2*9.8*8 = \sqrt156.96$

Hence, the value of v is 12.53m/s

For more information, refer to the link:-

https://brainly.com/question/19247046