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Chapter 6: Problem 23

Emil is tossing an orange of mass \(0.30 \mathrm{kg}\) into the air. (a) Emil throws the orange straight up and then catches it, throwing and catching it at the same point in space. What is the change in the potential energy of the orange during its trajectory? Ignore air resistance. (b) Emil throws the orange straight up, starting \(1.0 \mathrm{m}\) above the ground. He fails to catch it. What is the change in the potential energy of the orange during this flight?

Short Answer

Expert verified
Answer: (a) The change in potential energy is 0 J. (b) The change in potential energy is 2.943 J.

Step by step solution

01

(a) Calculate the change in potential energy when the orange returns to the initial point.

Since the orange returns to the same point in space, the change in height is \(0 \mathrm{m}\). So the change in potential energy is: \(\Delta PE = mgh = (0.3 \mathrm{kg})(9.81 \mathrm{m/s^2})(0 \mathrm{m}) = 0 \mathrm{J}\).
02

(a) Final answer for part (a)

There is no change in potential energy in this scenario, so \(\Delta PE = 0 \mathrm{J}\).
03

(b) Calculate the change in height from when Emil released the orange to when it hit the ground.

The orange is initially \(1.0 \mathrm{m}\) above the ground and falls to the ground. Therefore, the change in height is: \(\Delta h = 1.0 \mathrm{m}\).
04

(b) Calculate the change in potential energy for this scenario.

Now that we have the change in height, we can find the change in potential energy as follows: \(\Delta PE = mgh = (0.3 \mathrm{kg})(9.81 \mathrm{m/s^2})(1.0 \mathrm{m}) = 2.943 \mathrm{J}\).
05

(b) Final answer for part (b)

The change in potential energy in this scenario is \(\Delta PE = 2.943 \mathrm{J}\).

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