2. An 950 kg car is traveling on a flat level road at 30 m/s when the driver sla

Question

2. An 950 kg car is traveling on a flat level road at 30 m/s when the driver slams on the brakes and skids in a straight line for 3 seconds (he does NOT stop, just slows). The coefficient of friction between the car and the road is 0.662665. a. What is the initial momentum of the car? b. What is the initial kinetic energy of the car? c. What is the frictional force stopping the car? d. What is the impulse applied to the car while it is skidding? SD8 23345 What is the car's momentum after skidding? e. f. What is the car's speed after skidding? g. What is the car's kinetic energy after skidding? h. Using energy, determine how far the cartraveled while it was skidding

Explanation / Answer

Given,

m = 950 kg ; vi = 30 m/s ; t = 3 s

a)Pi = m vi

Pi = 950 x 30 = 28500 kg-m/s

Hence, Pi = 28500 kg-m/s

b)KEi = 1/2 m vi^2

KEi = 0.5 x 950 x 30^2 = 427500 J

Hence, KEi = 427500 J

c)Ff = uk N

Ff = 0.662665 x 950 x 9.81 = 6175.7064675 N

Hence, Ff = 6175.7064675 N

d)J = F t

J = 6175.7064675 x 3 = 18527.1194025

Hence, J = 18527.1194025 N s

e)since the momentum of the system is conserved, it will have the same value as before.

f)distance covered in 3 sec is:

d = 30/3 = 10 m

Wf = 6175.7064675 x 10 = 61757.064675 J

Now the car has:

KE = 427500 - 61757.064675 = 365742.935325 J

1/2 m vf^2 = 365742.935325

vf = sqrt (2 x 365742.935325/950) = 27.75 m/s

Hence, vf = 27.75 m/s

g)KEf = 365742.935325 J

h)KEi = Wf

6175.7064675 D = 427500

D = 69.2 m

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