Work-Engery Theorem?

Question

A big car of mass 2m travels at speed v, and a small car of mass m travels with a speed 2v. Both skid to a stop with same coefficient of friction. What is the ratio of the stopping distance of the small car to that of the large car. (Please help, Using Work-Engergy Theorem)

Answer 1

KE = W; where KE is kinetic energy of the cars when the brakes are engaged. W is the work of the friction force applied to the wheels of the cars as they skid to a halt. They equal because the work needed is to reduce the kinetic energy of each car to zero.

ke = 1/2 2mv^2 and KE = 1/2 m(2v)^2 the kinetic energies for the big and small cars respectively.

w = k(2mg)d and W = kmgD the work functions for the big (2m) car and small (m) cars respectively. d is the distance needed to work off ke from the large car and D is the distance to work off the KE from the small car.

Thus we have ke/KE = 1/2 2mv^2//1/2 4mv^2 = 1/2 = k(2mg)d/kmgD = 2d/D; so that d/D = 1/4 or,as you asked for the small car to large ratio D/d = 4; thus, the small car with the greater velocity will go four times farther than the large car.

Answer 2

The kinetic energy of an object is
E = 1/2 * m * v^2

The big car has mass 2m, so if everything else was the same, it would have twice the kinetic energy of the small car. However, all else is not the same. The small car has twice the velocity.

In total, the small car has 2^2/2 = 2 times as much kinetic energy as the big car.

The work done by friction is:
W = F*x
where F is the force and x is the distance over which that force is applied.

To bring each car to a stop, the frictional force must do an amount of work on the car to reduce its kinetic energy to zero. The work-energy theorem tells us that W = E.

The friction force is
F = mu*N
where mu is the coefficient of friction, and N is the normal load on the tires. N = m*g, where m is mass and g is gravity.

The bigger car can brake with twice the frictional force, due to its greater normal load. We know the small car has twice the kinetic energy of the big car, but the frictional force on the big car is twice as great, so we can say:

Small car:
2*E = 2*W = F * x

Big car:
E = W = 2*F * x

It becomes clear that for the small car, x will have 4 times the value it does for the large car. The small car will stop in 4 times the distance.

However, I must point out that F = mu*N is not accurate for real tires. At some point, increasing N will actually cause mu to decrease, leading to reduced traction. But for two normal vehicles, it's marginally accurate. Also notice that the size of the vehicle does not affect its stopping distance in our model, since the larger kinetic energy of a big vehicle is balanced precisely by its larger braking power due to the larger normal load. The only thing that affects stopping distance is velocity, and the distance goes as the velocity squared.

Answer 3

KE1 = 0.5 (2M)(V)^2 = MV^2

KE2 = 0.5 (M)(2V)^2 = 0.5M 4V^2 = 2MV^2

Your ratio = small car / big car