Newtons laws of force/ kinetic energy?

Question

Police investigators, examining the scene of an accident involving two cars, measure 72m long skid marks of one of the cars, which nearly came to a stop before colliding. The coefficent of kinetic friction between rubber and the pavement is about .80. estimate the initial speed of that car assuming a level road?

Answer 1

regular........


all the kinetic energy at the begining of the braking was consumed by the friction of the pavement in order to bring the car to a stop
(it is the 3rd time i write this today)


so

friction is Fr = n*N*= n*m*g so work is Fr*S
and kinetic energy is 1/2*m*u^2

therefore,

u = sqrt(2*n*g*s) =>

u=33.94m/s

Answer 2

Since the car skidded for 72 m before stopping,only friction is the force which decelerated the car.
Let 'k' be the coefficient of friction. Then the frictional force is 'kmg'.

So, Deceleration is kmg/m = 'kg'

Now applying the equation of motion :--

v^2 -u^2 = 2aS

we get

u = sqrt(2aS) = sqrt (2*0.8*10*72) = 33.8 m/s^2

Answer 3

This problem involves kinetic energy, force of kinetic friction, work energy, and conservation of energy.

The kinetic energy of the car=1/2mv^2, where m is its mass in kg and v its velocity before it screeched to a stop (nearly) before colliding.

This energy is converted to work energy, which is =f*s, where f is the force of kinetic friction in N, and s the displacement of 72m.

f=uR, where u is the coefficient of kinetic friction of 0.8, and R the force normal to the road, in this case the weight of the car. But we know that W=mg, where g=9.8m/s^2.

Now substitute known values:

f=0.8*mg
Work=f*s=0.8*mg*72

Based on the principle of conservation of energy:

KE=Work
1/2mv^2=0.8mg*72 (m cancels out, right?)
v^2=2*0.8*9.8*72
=1129
v =33.6m/s

Answer 4

Leave it for Sherlock Holmes...he will find it...if he can not...ask 007 James Bond...he will certainly find out...for you.