First slide

Work Energy Theorem

Question

A car is moving on a straight horizontal road with a speed v. If the coefficient of friction between the tires and the road is $μ$ , the shortest distance in which the car can be stopped is

Moderate

A

$V^{2} / 2 μg$
B

$v^{2} / μg$
C

$(v / μg)^{2}$
D

$v^{2} / μ$

Solution

$Initial K.E. of car = \frac{1}{2} {mv}^{2} work done against friction = μmgs Applying w o r k e n e r g y t h e o r e m y, we have μmgs = \frac{1}{2} {mv}^{2} or s = v^{2} / (2 μg)$

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