------------------------------------ coef. static friction ---------------------

Question

------------------------------------ coef. static friction -------------------------------- coef. kinetic friction
tires on dry road------------------------- 1.00 --------------------------------------------------0.800
tires on wet road----------------------- 0.600 -------------------------------------------------0.400
tires on snowy road------------------- 0.300 -------------------------------------------------0.200

Using the table above, the minimum stopping distance for a car of mass 2500.0 kg initially traveling at 20.0 m/s on a dry road will be meters.

Please show solution, thank you.

Explanation / Answer

We can solve this question by work energy principle.

The initial Kinetic energy(K.E) of the body is converted to work done by the friction.

initial K.E = 0.5*mv^2

Work done by friction = Friction*minimum stopping distance = F*x

x = minimum stopping distance

Friction force, F = umg

u = coefficient of kinetic friction on dry road = 0.8

m = mass = 2500

g = 20 m/s

So, 0.5*mv^2 = F*x = u*mg*x

So, x = v^2/ug = 20^2/(0.8*9.8) = 51.02 m <------------answer

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