2a. All KE was converted to PE during braking.
2b. a = (V^2-Vo^2)/2d.
a = (0-(5)^2) / 40 = -0.625 m/s^2.
Wc = mg = 400kg * 9.8N/kg = 3920 N. = Wt. of car
Fc = 3920N @ 0 Deg. = Force of car.
Fp = 3920*sin(0) = 0 = Force parallel to road.
Fv = 3920*cos(0) = 3920N. = Force perpendicular to road.
Fn = Fp - Fk = ma.
0 - Fk = 400*(-0.625).
Fk = 250 N. = Force of kineti friction.
2) A car traveling with 5 m/s encounters a deer. The drivers slams on the breaks, the breaks lock. The car
skids to a complete stop on the road ahead (and the deer runs away). Why does the car stop (where does
all the kinetic energy it had go)? Find the magnitude of the friction force if the skid marks show a 20 m slide
and the mass of the car mass is 400 kg.
Here
1 answer
1 answer