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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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Most popular questions from this chapter

A \(62-\mathrm{kg}\) woman takes \(6.0 \mathrm{s}\) to run up a flight of stairs. The landing at the top of the stairs is \(5.0 \mathrm{m}\) above her starting place. (a) What is the woman's average power output while she is running? (b) Would that be equal to her average power input-the rate at which chemical energy in food or stored fat is used? Why or why not?
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Justin moves a desk \(5.0 \mathrm{m}\) across a level floor by pushing on it with a constant horizontal force of \(340 \mathrm{N}\). (It slides for a negligibly small distance before coming to a stop when the force is removed.) Then, changing his mind, he moves it back to its starting point, again by pushing with a constant force of \(340 \mathrm{N}\). (a) What is the change in the desk's gravitational potential energy during the round-trip? (b) How much work has Justin done on the desk? (c) If the work done by Justin is not equal to the change in gravitational potential energy of the desk, then where has the energy gone?
Hilda holds a gardening book of weight \(10 \mathrm{N}\) at a height of $1.0 \mathrm{m}\( above her patio for \)50 \mathrm{s}$. How much work does she do on the book during that 50 s?
Two springs with spring constants \(k_{1}\) and \(k_{2}\) are connected in series. (a) What is the effective spring constant of the combination? (b) If a hanging object attached to the combination is displaced by \(4.0 \mathrm{cm}\) from the relaxed position, what is the potential energy stored in the spring for \(k_{1}=5.0 \mathrm{N} / \mathrm{cm}\) and $k_{2}=3.0 \mathrm{N} / \mathrm{cm} ?$ [See Problem \(83(\mathrm{a}) .]\)
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