Question

A 650-kg roller coaster starts from rest at the top of a 80 m hill.

a) What is the rollercoaster's Gravitational Potential Energy at the top
of the hill?

b) Assuming no friction, what is the rollercoaster's Kinetic Energy when it gets to the bottom of the hill?

c) Assuming no friction, what is the rollercoaster's speed when it gets to the bottom of the hill?

Answer 1

A. The the rollercoaster's Gravitational potential energy is 509600 J

B. The rollercoaster's kinetic energy is 509600 J

C. The rollercoaster's speed is 39.6 m/s

A. How to determine the potential energy

From the question given above,

• Mass (m) = 650 Kg
• Height (h) = 80 m
• Acceleration due to gravity (g) = 9.8 m/s²
• Potential energy (PE) =?

PE = mgh

PE = 650 × 9.8 × 80

PE = 509600 J

B. How to determine the kinetic energy

• Potential energy (PE) = 509600 J
• Kinetic energy (KE) =?

Since the rollercoaster is descending from the hill, the potential energy at the top will be equal to the kinetic energy at the bottom.

Thus,

KE = PE = 509600 J

C. How to determine the velocity

• Kinetic energy (KE) = 509600 J
• Mass (m) = 650 Kg

• Velocity (v) =?

KE = ½mv²

509600 = ½ × 650 × v²

509600 = 325 × v²

Divide both side by 325

v² = 509600 / 325

Take the square root of both side

v = √(509600 / 325)

v = 39.6 m/s

Learn more about energy:

https://brainly.com/question/10703928

#SPJ1