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Chapter 6: Problem 17

A 69.0 -kg short-track ice skater is racing at a speed of $11.0 \mathrm{m} / \mathrm{s}$ when he falls down and slides across the ice into a padded wall that brings him to rest. Assuming that he doesn't lose any speed during the fall or while sliding across the ice, how much work is done by the wall while stopping the ice skater?

Short Answer

Expert verified
Answer: The work done by the padded wall to stop the ice skater is 4174.5 Joules.

Step by step solution

01

Find the initial kinetic energy of the skater

First, we need to determine the initial kinetic energy of the skater before hitting the wall. The formula for kinetic energy is given as: KE = \(\frac{1}{2}mv^{2}\) where m is the mass and v is the velocity of the object. We can plug the given values into the formula: m = 69.0 kg (mass of skater) v = 11.0 m/s (velocity of skater)
02

Calculate the initial kinetic energy

Now we can calculate the initial kinetic energy: KE = \(\frac{1}{2}(69.0 \mathrm{kg})(11.0 \mathrm{m/s})^{2}\)
03

Perform the calculation

Calculating the initial kinetic energy, we get: KE = \(\frac{1}{2}(69.0)(121.0) = 4174.5 \mathrm{J}\) (Joules)
04

Determine the final kinetic energy

Since the skater comes to a complete stop after hitting the wall, his final kinetic energy is 0J.
05

Apply the work-energy theorem

Now that we have the initial and final kinetic energies, we can use the work-energy theorem to determine the work done by the wall: Work = Change in Kinetic Energy = Final KE - Initial KE Work = \(0 \mathrm{J} - 4174.5 \mathrm{J}\)
06

Calculate the work done

Now we can calculate the work done by the wall: Work = -4174.5 J The negative sign indicates that the work done is in the opposite direction of the initial movement of the skater. So, the work done by the padded wall to stop the ice skater is 4174.5 Joules.

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Most popular questions from this chapter

An airline executive decides to economize by reducing the amount of fuel required for long-distance flights. He orders the ground crew to remove the paint from the outer surface of each plane. The paint removed from a single plane has a mass of approximately \(100 \mathrm{kg} .\) (a) If the airplane cruises at an altitude of \(12000 \mathrm{m},\) how much energy is saved in not having to lift the paint to that altitude? (b) How much energy is saved by not having to move that amount of paint from rest to a cruising speed of $250 \mathrm{m} / \mathrm{s} ?$
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