Question

A roller coaster cart of mass 205.0 kg is pushed against a launcher spring with spring constant 600.0 N/m compressing it by 7.5 m in the process. When the roller coaster is released from rest the spring pushes it along the track (assume no friction in cart bearings or axles and no rolling friction between wheels and rail). The roller coaster then encounters a series of curved inclines and declines and eventually comes to a horizontal section where it has a velocity 6.0 m/s. How far above or below (vertical displacement) the starting level is this second (flat) level? If lower include a negative sign with the magnitude.​

Answer 1

First, let's calculate the potential energy stored in the spring:

PE = (1/2)kx²

where k is the spring constant and x is the distance the spring is compressed. Plugging in the given values, we get:

PE = (1/2)(600.0 N/m)(7.5 m)² = 16875 J

This potential energy is converted into kinetic energy as the roller coaster moves along the track. At the horizontal section, all of the potential energy has been converted into kinetic energy:

KE = (1/2)mv²

where m is the mass of the roller coaster and v is its velocity. Plugging in the given values, we get:

16875 J = (1/2)(205.0 kg)(6.0 m/s)²

Simplifying and solving for the vertical displacement, we get:

Δy = (KE/mg) - 7.5 m

where g is the acceleration due to gravity. Plugging in the values, we get:

Δy = [(1/2)(205.0 kg)(6.0 m/s)²/(205.0 kg)(9.81 m/s²)] - 7.5 m

Δy = 8.47 m

Therefore, the roller coaster is 8.47 meters above the starting level at the second (flat) level.

Step-by-step explanation: