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Chapter 6: Problem 21

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?

Short Answer

Expert verified
Answer: The change in the desk's gravitational potential energy during the round-trip is 0 J. The work done by Justin on the desk is 3400 J. The energy difference (3400 J) has gone into other forms of energy like heat due to friction between the desk and the floor.

Step by step solution

01

Identify the given information

The given information in the problem is: - Force applied, \(F = 340 \mathrm{N}\) - Distance moved, \(d = 5 \mathrm{m}\) - Gravitational potential energy change is 0 since the floor is level.
02

Calculate the work done by Justin in moving the desk

To calculate the work done, we use the formula: \(W = Fd\cos\theta\) Since the force is horizontal and the movement is also horizontal, the angle between them is 0 degrees. So, \(\cos\theta = \cos(0) = 1 \). Therefore, the formula becomes: \(W = Fd\) We can plug in the given values to find the work done for one movement (either forward or backward): \(W = 340 \mathrm{N} \times 5.0 \mathrm{m} = 1700 \mathrm{J}\)
03

Calculate the round-trip work done by Justin

Since the work done on the desk is the same for both movements (forward and backward), we multiply the single movement work done by 2 to find the total work done: Total Work Done \(= 2 \times 1700 \mathrm{J} = 3400 \mathrm{J}\)
04

Determine the change in the desk's gravitational potential energy

Since the movement is horizontal and the floor is level, there is no change in the desk's height. Therefore, there is no change in the gravitational potential energy. So, the change in the desk's gravitational potential energy is 0 J.
05

Answer the questions

(a) The change in the desk's gravitational potential energy during the round-trip is 0 J. (b) Justin has done 3400 J of work on the desk during the round-trip. (c) Since the work done by Justin (3400 J) is not equal to the change in gravitational potential energy (0 J), the difference in energy has gone into other forms of energy like heat due to friction between the desk and the floor.

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Most popular questions from this chapter

The bungee jumper of Example 6.4 made a jump into the Gorge du Verdon in southern France from a platform \(182 \mathrm{m}\) above the bottom of the gorge. The jumper weighed \(780 \mathrm{N}\) and came within \(68 \mathrm{m}\) of the bottom of the gorge. The cord's unstretched length is \(30.0 \mathrm{m}\) (a) Assuming that the bungee cord follows Hooke's law when it stretches, find its spring constant. [Hint: The cord does not begin to stretch until the jumper has fallen \(30.0 \mathrm{m} .]\) (b) At what speed is the jumper falling when he reaches a height of \(92 \mathrm{m}\) above the bottom of the gorge?
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