Physics prob. Urgent. Need Help. Deadline Tonight?

Question

You testify as an "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill (Fig. 6-27). You find that the slope of the hill is = 12.0°, that the cars were separated by distance d = 26.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 17.5 m/s.

At what speed did car a hit car b if coefficient of kinetic friction is .6?

and then if it is .1?


I've been at this problem for two days now. countless people have tried to hint at it, but i just don't understand it. An explanation w/ an answer would be great.

I have an 11 Pm deadline, or else I don't get credit.

Answer 1

Eng_helper was on the right track, but...

because of the slope, the normal force used to calculate the force of friction is less than the weight, W, of the car A. That normal force (straight into the road) is given by (W x cos 12 degrees.) or (0.97815 x W).

(When the angle is 0, the cos=1, so the normal force = the weight, as the angle goes up the cos goes down, and there is less and less normal force, the force perpendicular to the road)

In the first case, the force of friction is 0.6 x (that normal force) or (0.6 x 0.97815 x W) (usually the normal force on a flat surface would just be the weight of the car)

Because weight is a force, and F=mg, W, the weight of the car, is mg, where g is the acceleration of gravity and m is the car's mass.

You can substitute mg for W in the Normal Force expression and get the normal force = 0.6 x 0.97815 x mg.

So the force of deceleration, as the Eng_helper suggested, F will = (the coef of kinetic friction) x (the Normal Force perpendicular to the road) IF WE WERE IGNORING GRAVITY DUE TO THE SLOPE.

(but not ignoring it to calculate the normal force)

F = 0.6 x 0.97815 x mg

So the acceleration (deceleration in this case) due to the normal force alone is F/m, or (-0.6 x 0.97815 x mg / m). The m's cancel, and we are left with (-0.6 x 0.97815 x g)

And in the second problem (-0.1 x 0.97815 x g)

Now the ACCELERATION due to the gravity force down the slope is the component of g FACING down the slope. Draw your force vector straight down due to g and you will see that it breaks into 2 components, one down the slope and one (the Normal Force) perpendicular to the slope.

The acceleration of g down the slope is (sin(12 degrees) x g) or (0.20791 x g)

Net acceleration in the first case is -.58689g + .20791g or -.37898g

Now you will have a formula for distance and acceleration and initial velocity to plug the numbers into.

In the second case, the acceleration is -0.0978g + 0.20791g or 0.1100g. The car is indeed accelerating because the road is so very slippery! Use your formulae again about initial velocity, final velocity, and acceleration and distance.

Hope that helps!

Good luck!

All this was VERY back of the envelope. Will look again later tonight!

Answer 2

If there were no slope, the acceleration would be -.6g or -.1g for the second case.
Because of the slope, you have another acceleration to add to car A, which is g*sin 12.0 deg = .2g

Net acceleration on car A is -.6g + .2g = -.4g
So you have a car traveling 26m with an initial speed of 17.5m/s and an acceleration of -.4g
Now do your math. If your result is negative, it means that the car actually stopped before hitting car B.

In the second case you have a net accleration of -.1g + .2g = .1g (positive) which means that even if car A is trying to brake, its speed is increasing because of the slope of the road (.1 as coeficient of friction means that the road is very slippery).