Question

A 100 kg roller coaster is pulled up the first hill to a height of 8 meters. The car has an initial velocity of 2 m/s at the top of the first hill, rolls down the hill, and then up a second hill. (g = 10 m/s2)

A.Frictional forces are negligible. If at the top of the second hill the car has twice its initial velocity, then what must be the height of the second hill?

B. Calculate the work done by friction supposing the car only makes it half way up the second hill where it comes to a complete stop.




C. Calculate the work done by the net external forces that moved the car from where it was at rest on the ground to the top of the first hill with its initial velocity. Frictional forces are negligible.

Answer 1

A. To solve for the height of the second hill, we can use conservation of energy. The roller coaster has potential energy at the top of the first hill, which is converted to kinetic energy as it rolls down the hill. The roller coaster then has potential energy again at the top of the second hill, which is again converted to kinetic energy as it rolls down the second hill. We can equate the potential energy at the top of the first hill to the kinetic energy at the top of the second hill to find the height of the second hill:

Potential energy at top of first hill = mgh1 = 100 kg * 10 m/s^2 * 8 m = 8000 J

Kinetic energy at top of second hill = (1/2) * m * v2^2 = (1/2) * 100 kg * (2 * 2 m/s)^2 = 800 J

Equating these, we get:

mgh2 = (1/2)mv2^2

100 kg * 10 m/s^2 * h2 = (1/2) * 100 kg * (2 * 2 m/s)^2

h2 = (1/2) * (2 m/s)^2 / 10 m/s^2 + 8 m

h2 = 0.2 m + 8 m

h2 = 8.2 m

Therefore, the height of the second hill must be 8.2 meters.

B. To calculate the work done by friction, we need to first calculate the distance traveled by the roller coaster before coming to a stop halfway up the second hill. We can use conservation of energy again to do this:

Potential energy at top of first hill = mgh1 = 100 kg * 10 m/s^2 * 8 m = 8000 J

Kinetic energy at top of first hill = (1/2)mv1^2 = (1/2) * 100 kg * (2 m/s)^2 = 200 J

Total energy at top of first hill = 8200 J

When the roller coaster reaches halfway up the second hill, it has no kinetic energy and some potential energy due to its height above the ground. Let's call the height of the roller coaster at this point h3. The potential energy at this point is:

Potential energy halfway up second hill = mgh3

The total energy at this point is the same as the total energy at the top of the first hill:

8200 J = mgh3 + work done by friction

We can solve for the work done by friction:

work done by friction = 8200 J - mgh3

work done by friction = 8200 J - 100 kg * 10 m/s^2 * (8 m + h3)

To find h3, we need to use the fact that the roller coaster comes to a stop halfway up the second hill, which means its final velocity is 0. We can use conservation of energy again to find the height of the roller coaster at this point:

Potential energy at top of first hill = mgh1 = 8000 J

Kinetic energy at top of first hill = (1/2)mv1^2 = 200 J

Potential energy halfway up second hill = mgh3

Total energy halfway up second hill = 8200 J - work done by friction

Setting these equal, we get:

8000 J + 200 J = mgh3 + (8200 J - mgh3 - work done by friction)

8200 J = 8200 J - work done by friction

work done by friction = 0

This means that the roller coaster