Work and Energy Question

[moose45]

The toy car, after attaining a maximum speed, skids to a stop on a rough section of the surface. Which of the following quantities must be measured in order to calculate the length of the skid?

I. The mass of the car.
II. The coefficient of kinetic friction between the road and the wheels.
III. The initial speed of the car.

A. I and II only
B. II and III only
C. I and III only
D. I, II, and III

They say the answer is B but I thought it was D. Don't you need the mass of the car to find the friction force?? Thanks!

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[loveoforganic]

Try working through the equations you need using variables only (plugging no numbers in). I'm pretty sure mass ends up cancelling. i.e. a 200 pound car going 30 miles per hour will stop in the same distance that a 2000 pound car going 30 miles per hour will. This is because the first one's lack of kinetic energy is compensated by the second car's increase in frictional force.

 

[BloodySurgeon]

Energy Conservation
W = F x d = delta KE
delta KE = 1/2mVf^2 - 1/2mVi^2
delta KE = 1/2mVi^2

F x d = 1/2mVi^2
Mkmgd = 1/2mVi^2
Mkgd = 1/2Vi^2 (NOTICE HOW THE MASS CROSSES OUT)

d = Vi^2 / 2gMk

Kinetics
Vf^2 = Vi^2 + 2ax
0 = Vi^2 + 2ax
Vi^2 = 2ax
a = Vi^2/2x

Mk*mg = ma
Mk*g = a (NOTICE HOW THE MASS CROSSES OUT)
Mk*g = Vi^2/2x

x = Vi^2 / 2gMk

Therefore the answer is B.

 

[moose45]

I worked it out:

Energy(initial) + Work Done = Energy(Final)

.5mv(initial)^2 - coefficient of friction.(mg).d = 0

.5mv(initial)^2 = coefficient of friction.(mg).d

d = v(initial)^2 / 2 coefficient of friction g

Is this correct when working out?

 

[moose45]

Thanks BloodySurgeon