Question

A 3.0 kg toy car is released at the top of a frictionless track on the left and rolls off of the track from its right side ramp. The car starts at a height of 0.80 m, goes through a 0.50 m diameter loop, and exits the ramp at a height of 0.25 m.

What is the change in the car’s gravitational potential energy from A to B?

Answer 1

A 3.0 kg toy car is released at the top of a frictionless track on the left and rolls off of the track from its right side ramp. The car starts at a height of 0.80 m, goes through a 0.50 m diameter loop, and exits the ramp at a height of 0.25 m.

What is the change in the car’s gravitational potential energy from A to B?

Answer: 7.35 J

Answer 2

ΔU= U2-U1= 7.35 J - 0 J= 7.35 J.

Step-by-step explanation:

Point A= 0 m.

Point B= 0.25 m.

Gravitational potential energy formula:
`U=mgh`, where m is the mass (in kg) of the body. G is the magnitude of the gravitational field that the body is being exposed to. H is the height at which the body is located at.

1. First, let's calculate gravitational potential energy of the car on point A:

U₁= (3.0 kg)(9.8 m/s²)(0 m)= 0 J.

2. Gravitational potential energy of the car on point B:

U₂= (3.0 kg)(9.8 m/s²)(0.25 m)= 7.35 J.

Now, calculating the change in the car’s gravitational potential energy:

ΔU= U₂-U₁= 7.35 J - 0 J= 7.35 J.