Physics ut onramps

Question

This question is typical on some driver's license exams: A car moving at 44 km/h skids

15 m with locked brakes.

How far will the car skid with locked brakes at 110 km/h? Assume that energy loss is due only to sliding friction.

Answer in units of m.

Answer 1

Let's use SI units. 44 km/h x 1000m/3600s = 110/9 m/s and 110 km/hr = 275/9 m/s

The initial energy (kinetic) = 1/2mv² = 1/2m(110/9)² = 74.69m joules

The final energy = 0

Then the work done is 74.69m joules (lost due to friction)

F_F = μN = μmg = 9.81μm

Work = Fd = (9.81μm)(15) = 147.15μm

So 147.15μm = 74.69m or 147.15μ = 74.69 meaning μ = 74.69/147.15 = 0.5071

If the car is traveling 275/9 m/s (110km/hr) then its kinetic energy = 1/2mv² = 1/2m(275/9)² = 466.82m joules

The work will be Fd = μNd = μmgd = (0.5071)m(9.81)d = 4.9794md

So 466.82m = 4.9794md or 466.82 = 4.9794d meaning d = 466.82/4.9794 = 93.75 m = 94 meters