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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 6 / Problem 29E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: Stopping Distance. A car is traveling on a level

Chapter 6, Problem 29E

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QUESTION:

Problem 29E

Stopping Distance. A car is traveling on a level road with speed v0 at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work–energy theorem to calculate the minimum stopping distance of the car in terms of v0, g, and the coefficient of kinetic friction k between the tires and the road. (b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

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QUESTION:

Problem 29E

Stopping Distance. A car is traveling on a level road with speed v0 at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work–energy theorem to calculate the minimum stopping distance of the car in terms of v0, g, and the coefficient of kinetic friction k between the tires and the road. (b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

ANSWER:

Solution 29E

Step 1:

Work energy theorem states

$W\ =\ \Delta{}KE=\ (1/2\ m{{v}_{f}}^{2})-(1/2\ m{{v}_{i}}^{2})$ ------------(1)

Work done can be written as $W=\ -{M}_{K}mgL\$

When the car reaches final point its velocity is ${v}_{f}=\ 0\$ hence $(1/2\ m{{v}_{f}}^{2})=\ 0\$

Thus we can write can as

$-{M}_{K}mgL=\ (1/2\ m{{v}_{i}}^{2})$

Thus we get L as $L=\ \frac{{{v}_{i}}^{2}}{2{M}_{K}g}$

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