A 2600 kg automobile is traveling at 20 m/s when the brakes are applied. The bra

Question

A 2600 kg automobile is traveling at 20 m/s when the brakes are applied. The braking force is 10192 N. Use work-energy methods for this problem.

a. Determine the initial kinetic energy of the automobile. joules

b. Determine the work required to stop the automobile. joules

c. Find the distance required to stop the car. m

Now, let's repeat this problem, but with the driver moving at two times the initial speed. That is, at 40 m/s

d. Determine the initial kinetic energy of the automobile. joules

e. Determine the work required to stop the automobile. joules

f. Find the distance required to stop the car. m


g. How do the stopping distances compare when the speed has been doubled?

Explanation / Answer

a) initial kinetic energy is (1/2)*m*(v^2) = (1/2)*2600*20*20 = 520000 J According to the work-energy theorem, if one or more external forces act upon a rigid object, causing its kinetic energy to change from Ek1 to Ek2, then the work (W) done by the net force is equal to the change in kinetic energy so for (b) work required to stop the automobile is = 520000 J as final kinetic energy of automobile = 0 initial = 520000 J 520000 - 0 = 520000 J C) decelaration of the car is 10 192 / 2600 = 3.92 m/s^2 applying the basic kinematic equation V^2 - U^2 = 2*a*s V =0 U = 20 ,a = -3.92 so s = 400/3.92 = 102.041 m (d) when the speed is doubled i.e. V` = 2*V then kinetic energy is = (1/2)*m*(v^2) = (1/2)*2600*40*40 = 2080000 J e) work required to stop is = 2 080 000 J f) distance required is decelaration of the car is 10 192 / 2600 = 3.92 m/s^2 applying the basic kinematic equation V^2 - U^2 = 2*a*s V =0 U = 40 ,a = -3.92 so s = 1600/3.92 = 408.164 m g) stopping distance become 4 times when speed is doubled

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