A roller coaster of mass 2000 kg is rolling down a track with an instantaneous speed of 10 m/s. The roller coaster has a force of
unknown magnitude act on it, so now it is traveling at only 6
m/s. What was the work done on it?

The work done by the external force on the roller coaster is 64000 J.

Given that the mass of The roller coaster is 2000 kg and its velocity is 10 m/s. The kinetic energy of the roller coaster is given below.

Since the roller coaster has a force of unknown magnitude which decreases its speed to 6 m/s. For this decreased speed, the kinetic energy is given as below.

The work done on the roller coaster is calculated by the change in the kinetic energy for both speeds.

Work Done = KE - KE'

Work = 100000 - 36000

Work = 64000 J

Hence we can conclude that the work done by the external force on the roller coaster is 64000 J.

To know more about the work, follow the link given below.

link.

Kinetic energy = (1/2) (mass) (speed)²

The coaster's original kinetic energy was

KE = (1/2) (2,000 kg) (10 m/s)²

KE = (1,000 kg) (100 m²/s²)

KE = 100,000 kg-m²/s²

old KE = 100,000 joules

After something slowed it down, its new kinetic energy is

KE = (1/2) (2,000 kg) (6 m/s)²

KE = (1,000 kg) (36 m²/s²)

KE =  36,000 kg-m²/s²

new KE = 36,000 joules

The coaster lost (100,000 - 36,000) = 64,000 joules of KE.

Energy is conserved.  It can't evaporate, disappear, or be destroyed.

That 64,000 joules had to go somewhere.

Someone or something had to do 64,000 joules of work on it in order to slow it down.

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