Question

1. Potential energy and a roller coaster. A 1000-kg roller coaster car moves from point A to B then C.

Answer 1

Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,

          U

B

​

=0

When the car is at point A, the potential energy of the car-Earth system is given by

          U

A

​

=mgy

where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:

y=(41.1m)sin40.0

0

=26.4m

Thus,

U

A

​

=(1000kg)(9.80m/s

2

)(26.4m)=2.59∗10

5

J

The change in potential energy of the car-Earth system as the car moves from A to B is

U

B

​

−U

A

​

=0−2.59∗10

5

J=−2.59∗10

5

J

(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U

A

​

=0. The potential energy of the system when the car is at point B is given by U

B

​

=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.

Thus,