Question

A woman is driving her truck with speed 45.0 mi/h on a horizontal stretch of road. (a) When the road is wet, the coefficient of static friction between the road and the tires is 0.105. Find the minimum stopping distance (in m). m (b) When the road is dry, μs = 0.602. Find the minimum stopping distance (in m).

Answer 1

Answer:

Explanation:

45 mi /h = 45 x 1.6 x 1000 / (60 x 60) m /s

= 20 m /s

Maximum frictional force possible on wet road

= μs x mg where μs is coefficient of static friction and m is mass of body

Applying work energy theorem

work done by friction = kinetic energy of truck

μs x mg x d  = 1/2 m v ² where v is velocity of body and d is stopping distance

2xμs x g x d = v²

2 x .105 x 9.8 x d = 20 x 20

d = 194.36 m

b )

In this case

μs = 0.602

Inserting this value in the relation above

2xμs x g x d = v²

2 x .602 x 9.8 x d = 20 x 20

d = 33.9 m .

Answer 2

When the road is wet, the minimum stopping distance is 196.7 m

When the road is dry, the minimum stopping distance is 34.3 m.

The given parameters;

• speed of the truck, v = 45 mi/h

The speed of the truck in m/s is calculated as follows;

$v = \frac{45 \ mi}{h} \times \frac{1609.34 \ m}{1 \ mile} \times \frac{1 \ h}{3600 \ s} \\\\v = 20.12 \ m/s$

The work done by friction is calculated by applying work-energy theorem;

$F_k d = \frac{1}{2} mv^2\\\\\mu_k mg d = \frac{1}{2} mv^2\\\\ 2 \mu_k g d = v^2\\\\d = \frac{v^2}{2\mu_k g}$

When the road is wet, the minimum stopping distance is calculated as follows;

$d = \frac{20.12^2}{2 \times 0.105 \times 9.8} \\\\d = 196.7 \ m$

When the road is dry, the minimum stopping distance is calculated as follows;

$d = \frac{(20.12)^2}{2\times 0.602 \times 9.8} \\\\d = 34.3 \ m$

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