Question

1. you are sitting in a cart at the top of a hill on a roller coaster. you have 41,200 j of gpe. if your current weight is 539 n, how tall is the hill you are sitting on?
2. what is the potential energy of your 12 kg puppy that is sitting in the grass in your backyard?
3. in a lab investigation, one group of students (group a) measures the speed of a 0.10 kg toy car to be 2.5 m/s at the bottom of a ramp. another group of students (group b) measures the speed of the car to be 3.0 m/s at the bottom of the ramp. the cars starting position at the top of the ramp is 1.5 m high.

a. what is the potential energy of the car at the beginning of the experiment before its speed is measured?
b. calculate the kinetic energy of the car for group a.
c. calculate the kinetic energy of the car for group b.

1. challenge: you are at the grand canyon, standing at the edge of a ledge 1,505 m high. you have a mass of 56.9 kg. you decide to take a selfie to share with your science teacher when you get home which causes you to wonder...

a. how much gravitational potential energy do you have standing at the edge of this cliff?
b. if you were to trip and fall mid - selfie, your stored potential energy would be converted to kinetic energy, as you are in motion. assume that all of the gpe you calculated in part a would be converted to kinetic energy when you fall. if so, how fast would you be falling?

Explanation:

Step1: Write the formula for gravitational potential energy

The formula for gravitational potential energy \(GPE = mgh\), where \(m\) is mass, \(g = 9.8\ m/s^{2}\) (acceleration due to gravity), and \(h\) is height. Also, weight \(W=mg\).

Step2: Solve for height in problem 5

Given \(GPE = 41200\ J\) and \(W = 539\ N\). Since \(GPE=W\times h\), then \(h=\frac{GPE}{W}\).
Substitute the values: \(h=\frac{41200}{539}\approx76.44\ m\)

Step3: Solve for potential energy in problem 6

If the puppy is sitting in the grass, \(h = 0\ m\). Using \(GPE=mgh\), with \(m = 12\ kg\), \(g = 9.8\ m/s^{2}\), and \(h = 0\ m\). Then \(GPE=12\times9.8\times0 = 0\ J\)

Step4: Solve for potential energy in problem 7a

Given \(m = 0.10\ kg\), \(h = 1.5\ m\), \(g = 9.8\ m/s^{2}\). Using \(GPE=mgh\), \(GPE=0.10\times9.8\times1.5=1.47\ J\)

Step5: Solve for kinetic energy in problem 7b

The formula for kinetic energy \(KE=\frac{1}{2}mv^{2}\). For group A, \(m = 0.10\ kg\), \(v = 2.5\ m/s\). Then \(KE=\frac{1}{2}\times0.10\times(2.5)^{2}=0.3125\ J\)

Step6: Solve for kinetic energy in problem 7c

For group B, \(m = 0.10\ kg\), \(v = 3.0\ m/s\). Using \(KE=\frac{1}{2}mv^{2}\), \(KE=\frac{1}{2}\times0.10\times(3.0)^{2}=0.45\ J\)

Step7: Solve for gravitational potential energy in problem 8a

Given \(m = 56.9\ kg\), \(h = 1505\ m\), \(g = 9.8\ m/s^{2}\). Using \(GPE=mgh\), \(GPE=56.9\times9.8\times1505 = 56.9\times14749=839218.1\ J\)

Step8: Solve for velocity in problem 8b

If \(GPE = KE\), and \(KE=\frac{1}{2}mv^{2}\), \(GPE=mgh\). Then \(mgh=\frac{1}{2}mv^{2}\). Cancel out \(m\) (since \(m
eq0\)), \(v=\sqrt{2gh}\)
Substitute \(g = 9.8\ m/s^{2}\), \(h = 1505\ m\), \(v=\sqrt{2\times9.8\times1505}=\sqrt{29598}\approx172.04\ m/s\)

Answer:

1. The height of the hill is approximately \(76.44\ m\)
2. The potential energy of the puppy is \(0\ J\)

7a. The potential energy of the car is \(1.47\ J\)
7b. The kinetic energy of the car for group A is \(0.3125\ J\)
7c. The kinetic energy of the car for group B is \(0.45\ J\)
8a. The gravitational potential energy is \(839218.1\ J\)
8b. The falling speed is approximately \(172.04\ m/s\)