Can you explain step by step , how this person solved this physics question . 1) The coefficient of kinetic friction bet

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Can you explain step by step , how this person solved this physics question . 1) The coefficient of kinetic friction bet

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Can you explain step by step , how this person solved
this physics question .
1) The coefficient of kinetic friction between tires and a
typical road (asphalt) is µk = 0.55. Depending on the tires as well
as road conditions, µk can vary by as much as 25%. Based on this
information, the defense lawyer of a speeding ticket case argues
that the prosecutor interpretation of the tire marks -- providing
the vehicle stoppage distance -- is unreliable. Determine if the
defense argument is valid on a horizontal as well as on an inclined
road (angle between road surface and horizontal direction is at
most 10°). Consider that the vehicle was in a 25mph speed limit
zone and the tires left a 15-meters long mark on the road .
Can You Explain Step By Step How This Person Solved This Physics Question 1 The Coefficient Of Kinetic Friction Bet 1
Can You Explain Step By Step How This Person Solved This Physics Question 1 The Coefficient Of Kinetic Friction Bet 1 (581.53 KiB) Viewed 25 times
Can You Explain Step By Step How This Person Solved This Physics Question 1 The Coefficient Of Kinetic Friction Bet 2
Can You Explain Step By Step How This Person Solved This Physics Question 1 The Coefficient Of Kinetic Friction Bet 2 (654.08 KiB) Viewed 25 times
-Here, llp=0.55 V 25 toph = 14.76 받 Head of = lamg. Decelosenlion (2)=-llag Ok, s = 124.093 Now, we W² = 2as org O- (1116) -- Zlags Ok, 124.903= Illags = 11.57m 2x0.55X9.81 For, 25% variation, eltermax=0.55+ 25 X0.55 = 0.6875 TOO u konn = 0.55-25 x0.55=0.4125 100 Now, Smase 9.26m Smin = 15.433m As, Swin = 15.433> 15m, the defenses oreguomend us Cardeerd, because, i l=0.4125 and, s=15m. then, V = Jalgs – J9X04195X8-81X15
ok, ve 11.02m/s <w=11. 176 m/s (speed Limit) From figure V mg Sing - ma Co2 9 - On inclination Cassume up, since mal mentionel specifically). , mg Sino + le mg Coso are, a= g g Sing+ llag Coso Figure If, l = 0.4125 ,O= 10° a= 9.81 Sin loo +0.4125X9.81 Cooloº az 5.689 me Now, v=has J215689)X15 Oreg V=13.06 m/s Also, ple=0.5897536-15 az 9.81 Sin 10° +0.6875 X9.81 Cosloo Ore ga = 8.345 m/s2 Nowy V 120 8 = Jac8345) (5) = (15 Okey &= 15.82 m/s Both speeds, 15.82 and 13.06 are more than speed limit, W'=11176 m/s. So, argument not valid SAW
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