At an accident scene on a level road, investigators measure a car's skid mark to be 89 m long. The accident occurred on a rainy day, and the coefficient of kinetic friction was estimated to be 0.44.
1) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.
2) Why does the car's mass not matter?
Please explain #2 thoroughly because I have no idea why.
1) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.
2) Why does the car's mass not matter?
Please explain #2 thoroughly because I have no idea why.
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Step:
1) Draw a free body diagram. There are 3 forces acting on the car. mg pointing down for the car's weight, normal force (N) pointing up, and kinetic friction (fk) pointing backwards.
2) Determine the x and y components of net force.
Fx: -fk=ma
Fy: N-mg=0
3) Solve for a.
From Fy, you can determine that N=mg. You know that fk=mu(N). Plug in N=mg to get fk=mu(mg). Now your Fx equation looks like -mu(mg)=ma. Since you have masses on both sides of the equation, you can cross out mass (explaining part two). That leaves you with a=-mug=-.44(9.8)=-4.312 m/s^2
4) Use motion equations to determine v. Since you don't know t, choose and equation which does not have t. One such equation is vf^2=vo^2+2ax. (vo is initial velocity, vf is final velocity) You know that final velocity is 0 because the car has come to a rest. x was given in the problem, and you already figured out a previously. So just plug in and solve for vo to get the answer for part 1.
vo^2=vf^2-2ax=0-2(-4.312)(89)=767.54
vo=sqrt(767.54)=27.7 m/s
Solved.
1) Draw a free body diagram. There are 3 forces acting on the car. mg pointing down for the car's weight, normal force (N) pointing up, and kinetic friction (fk) pointing backwards.
2) Determine the x and y components of net force.
Fx: -fk=ma
Fy: N-mg=0
3) Solve for a.
From Fy, you can determine that N=mg. You know that fk=mu(N). Plug in N=mg to get fk=mu(mg). Now your Fx equation looks like -mu(mg)=ma. Since you have masses on both sides of the equation, you can cross out mass (explaining part two). That leaves you with a=-mug=-.44(9.8)=-4.312 m/s^2
4) Use motion equations to determine v. Since you don't know t, choose and equation which does not have t. One such equation is vf^2=vo^2+2ax. (vo is initial velocity, vf is final velocity) You know that final velocity is 0 because the car has come to a rest. x was given in the problem, and you already figured out a previously. So just plug in and solve for vo to get the answer for part 1.
vo^2=vf^2-2ax=0-2(-4.312)(89)=767.54
vo=sqrt(767.54)=27.7 m/s
Solved.
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Hi these questions are old news these days with ABS braking systems as they usually leave no skid mark any more what you might see however are marks suggesting a collision due to bits of the car along a section of road from where the collision occur ed.