Okay the question in the book says this... A care is traveling on a level road w

Question

Okay the question in the book says this...
A care is traveling on a level road with a speed of Voat the instant when the breaks lock, so that the tires slide ratherthan roll. (a.) Use the work-energy theorem to calculate the minimumstopping distance of the car in terms of Vo, g, and the coefficientof friction Uk between the tires and the road. (b.) By what factor would the minimum stopping distance changeif (i) the coefficient of kinetic friction were doubled, or (ii)the initial speed were doubled, or (iii) both the coefficient ofkinetic friction and the initial speed were doubled?
My question for homework is asking...
If your car is traveling at 78 km/hr and stops in 63.5 meters,what is the stopping distance if the car is traveling 61 km/hr?Give your answer in meters to the first decimal place.
The computer system says the answer = 38.8 meters. (I need toknow the steps needed to obtain this answer, will give greatratings for clear step by step responses. Thanks.)
A care is traveling on a level road with a speed of Voat the instant when the breaks lock, so that the tires slide ratherthan roll. (a.) Use the work-energy theorem to calculate the minimumstopping distance of the car in terms of Vo, g, and the coefficientof friction Uk between the tires and the road. (b.) By what factor would the minimum stopping distance changeif (i) the coefficient of kinetic friction were doubled, or (ii)the initial speed were doubled, or (iii) both the coefficient ofkinetic friction and the initial speed were doubled?
My question for homework is asking...
If your car is traveling at 78 km/hr and stops in 63.5 meters,what is the stopping distance if the car is traveling 61 km/hr?Give your answer in meters to the first decimal place.
The computer system says the answer = 38.8 meters. (I need toknow the steps needed to obtain this answer, will give greatratings for clear step by step responses. Thanks.)

Explanation / Answer

i) 1/2
ii) 4
iii) 2

It's done this way:

E = (1/2)mV^2 = FS

Where;
E = energy
m = mass
V = velocity
S = stopping distance
F = force of friction

if F = Fo; S = So ; V = Vo

where;
Fo = the normal or original force of friction
So = the normal or original stopping distance

Thus;

So = (1/2)m(Vo)^2/Fo

i) If the F = 2Fo

S = (1/2)m(Vo)^2/(2Fo)

Hence;

k = S/So = 1/2; where k is the factor

ii) If V = 2Vo

S = (1/2)m(2Vo)^2/Fo = 4(1/2)m(Vo)^2/Fo

k = S/So = 4

iii) If V=2Vo and F = 2Fo

S = 4(1/2)m(Vo)^2/(2Fo) = 2(1/2)m(Vo)^2/Fo

k = S/So = 2

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