Question
(a) If the coefficient of kinetic friction between tires and dry pavement is $0.80,$ what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 $\mathrm{m} / \mathrm{s}$ (about 65 $\mathrm{mi} / \mathrm{h} ) ?$ (b) On wet pavement the coefficient of kinetic friction may be only $0.25 .$ How fast should you drive on wet pavement in order to be able to stop inthe same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)
Step 1
In this case, the normal force is equal to the weight of the car, so $N=mg$ where $m$ is the mass of the car and $g$ is the acceleration due to gravity. Therefore, the force of kinetic friction is $f_{k}=\mu_{k} mg$. Show more…
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(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop a car by locking the brakes when the car is traveling at 28.7 m/s (about 65 mi/h)? (b) On wet pavement the coefficient of kinetic friction may be only 0.25. How fast should you drive on wet pavement to be able to stop in the same distance as in part (a)? ($Note$: Locking the brakes is $not$ the safest way to stop.)
Applying Newton's Laws
Friction Forces
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