a motorist in a car travelling with constant linear velocity brakes suddenly on a wet road and the car continu

Question

a motorist in a car travelling with constant linear velocity brakes suddenly on a wet road and the car continues to slide forward in the same direction. by considering the total kenetic energ;y of the wheels of the car before and after it skids and assuming that there is little fric tion between the wheel and ther road. explain whether the velocity of the car should increase when the brakes are applied

Answer 1

There should be no increase in velocity. Assuming no friction, an increase in velocity would require an increase in force because it is an acceleration. There is no increase in force/acceleration thus no increase in velocity.

Answer 2

You are trying to lay the groundwork for the argument that the rotational kinetic energy of the tires must be added to the linear KE of the vehicle. There is no mechanism for that transfer because you are supposing a frictionless slide.

Therefore, the net forces acting on the car from outside must be zero. There is nothing that can accelerate the car.

Also, if you are attempting to use conservation of angular momentum, then you have to pick an axis of rotation to conserve about. Any axis of rotation that shows a non-zero momentum for the rotating tires (i.e., any axis that is parrallel to the axles of the car) is going to show a net angular momentum of zero as the actual car itself isn't rotating about any axis parallel to the axles.

So, the car WILL NOT speed up because the brakes were applied.

Answer 3

On a level road, refer to answer #3. On a steep incline, it gets much more complicated - depending on the rolling resistance before the brakes are applied, and the friction when they are locked up.

Answer 4

The speed wouldn't increase, it would slow down from the friction of the water, assuming no force was being applied.

Answer 5

remeber the conservation of energy:

K(final) - K (initial) = U (final) - U (initial)

or let's write it this way and include friction in the formula

consider that there is no potential differnce so

k(final) = K(initial) + work done by friction

work done by friction is negative because the direction of friction is opposite of the displacement and so angle = 180 and sin 180 = -1

k(final) = K(initial) - work done by friction

delta k = - work done by friction

0.5 mv(final)^2 - 0.5mv(inital)^2 = work done by friction

or

0.5 mv(final)^2 - 0.5mv(inital)^2 = a negative value

so v(final) < v(initial)

so the speed of car decreases