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Physics / College Physics 1 / Chapter 7 / Problem 19

College Physics | 1st Edition | ISBN: 9781938168000 | Authors: Paul Peter Urone, Roger Hinrichs

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Solved: In Example 7.7, we found that the speed of a

Chapter 7, Problem 19

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QUESTION:

Problem 19PE

In Example 7.7, we found that the speed of a roller coaster that had descended 20.0 m was only slightly greater when it had an initial speed of 5.00 m/s than when it started from rest. This implies that ΔPE >> KEi . Confirm this statement by taking the ratio of ΔPE to KEi . (Note that mass cancels.)

Example 7.7 Finding the Speed of a Roller Coaster from its Height

(a) What is the final speed of the roller coaster shown in Figure 7.8 if it starts from rest at the top of the 20.0 m hill and work done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m/s?

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QUESTION:

Problem 19PE

In Example 7.7, we found that the speed of a roller coaster that had descended 20.0 m was only slightly greater when it had an initial speed of 5.00 m/s than when it started from rest. This implies that ΔPE >> KEi . Confirm this statement by taking the ratio of ΔPE to KEi . (Note that mass cancels.)

Example 7.7 Finding the Speed of a Roller Coaster from its Height

(a) What is the final speed of the roller coaster shown in Figure 7.8 if it starts from rest at the top of the 20.0 m hill and work done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m/s?

ANSWER:

Solution 19PE

When a body of mass $m$ is dropped from a certain height, say $h$ , the change in its potential energy is given by $\Delta{}PE=mgh$ . The expression for kinetic energy of an object having mass $m$ and velocity $v$ is given by $KE=\frac{1}{2}m{v}^{2}$ . We are required to show that in the example given, the change in potential energy is much greater than the initial kinetic energy.

Step 1 of 3

The roller coaster has descended by 20.0 m.

Let its mass be $m$ .

Therefore, the change in potential energy $\Delta{}PE=mg\times{}20.0$ J

$=m\times{}9.80\times{}20.0$ J

$=196m$

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