If the coefficient of kinetic friction between tires and dry pavement is 0.75, w

Question

If the coefficient of kinetic friction between tires and dry pavement is 0.75, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 35.0 m/s^2?
d=_______m

On wet pavement, the coefficient of kinetic friction may be only 0.25. How fast should you drive on wet pavement in order to be able to stop in the same distance as in part A? (Note: Locking the brakes is not the safest way to stop.)
v=_______m/s


show work if possible please!

Explanation / Answer

The kinetic energy is lowered by this work due to friction, which is force times path length:: ?E_kin = F_f · ?s Frictional force is frictional coefficient times normal force, which is equal to the weight of the car: F_f = µ·m·g Hence: ?(m·v²/2) = µ·m·g·?s ?(v²/2) = µ·g·?s Therefore the distance it takes to come at rest is: ?s = v²/(2· µ·g) = (25m/s)² / (2 · 0.8 · 9.81m/s²)) = 39.82m For wet pavement: ?s_wet = v_wet²/(2· µ_wet·g) The stopping is the same as for dry pavement: ?s_wet = ?s_dry =v_dry²/(2· µ_dry·g) => v_wet²/(2· µ_wet·g) = v_dry²/(2· µ_dry·g) v_wet = v_dry · v(µ_wet/µ_dry) = 25m/s · v(0.25 / 0.8) = 13.98m/s

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