Wc = M*g = 1300 * 9.8 = 12,740 N. = Normal force, Fn.
Fp = Mg*sin A = Mg*sin 0 = 0. = Force parallel to the plane.
Fk = u*Fn = 0.8 * 12,740 = 10,192 N.
a = (Fp-Fk)/m = (0-10,192)/1300 = -7.84 m/s^2.
d = (Vf^2-Vo^2)/2a = (0-(28^2))/-15.68 = 50 m.
A car of mass 1300 kg is driving along a road at speed of 28m/s. As it approaches some traffic lights, the lights change to red and the driver applies his brakes causing the car to skid. The frictional force between the skidding tyres and the road is 0.8 of the weight of the car. Caculate how far the car will travel before it stops?
I have calculated the distance to be 199.796m S=(V^2-V^2)/2a
& a = (F-X)/m = -1.962m/s/s
Are my calculations correct?
Any help would be greatly appreciated.
Best regards Sean
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