A 2500kg truck traveling at 30mph is sliding down a smooth ice covered road and

Question

A 2500kg truck traveling at 30mph is sliding down a smooth ice covered road and travels a distance of 1000m on the incline (30° relative to the horizontal). If it travels 450m up a runaway ramp inclined at an angle of 45° and covered with loose stone, calculate the effective coefficient of friction of the stones.

I know that this problem should be divided into 2 parts, and I'm having problems getting the correct final velocity for the first part, so in turn, can not get the correct coefficient of friction. (I already know that the coefficient of friction should be =0.6)

Explanation / Answer

Given Mass of the truck, m = 2500 kg Initial velocity of the truck on the smooth ice is, v = 30 mph = 13.4 m/s Distance travelled , s = 1000 m Angle of incline for the ice road, 1 = 30^0 Angle of incline for the stones road, 2 = 45^0 Distance travelled on the stone road, d = 450 m Net force acting on the truck on the ice road is, ma = mg sin30 a = (9.8 m/s^2) sin30= 4.9 m/s^2 The velocity of the truck at the end of the ice road is, vf^2 = v^2 + 2 as        = (13.4 m/s)^2 + 2(4.9 m/s^2)(1000 m) vf = 98.99 m/s ------------------------------------------------------------------------ The truck travels a runway with a speed of vf , now the work done by the net force acting on the truck is, w = 1/2 m vf^2 (mg sin 45 + mg cos45)*d =  1/2 m vf^2 = (vf^2 / 900 g cos45) - g sin 45    = ((98.99 m/s)^2 /900 (9.8 m/s^2) cos45) - 9.8 m/s^2 sin 45 = 0.57 ˜ 0.6

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