1)In an experiment on stopping distance, a truck drives 60mph, then the brakes a
ID: 1678979 • Letter: 1
Question
1)In an experiment on stopping distance, a truck drives 60mph, then the brakes are put on hard, so that the truck's wheelslock, and the truck skids to a stop. When the truck is empty, ithas a mass of 3000 kg, and it skids to a stop in 30 m. Whatdistance (in m) does the truck skid when it is loaded with a2,027-kg mass? (Ans: 3.00x10^1) 2) The terminal velocity of a 0.00008-kg drop is 1.5 m/s.Assuming a drag force F_D = -bv, what is the value of the constantb (in kg/s)? (Ans: 5.23x10^ -4) 3) The drag force on large objects moving through air is morenearly F_D = -bv^2. For this quadratic dependence on v, what is thevalue of the constant b (in Ns^2/m^2) if a 79-kg sky diver has aterminal velocity of 51 m/s? (Ans: 2.98*10^-1 )1)In an experiment on stopping distance, a truck drives 60mph, then the brakes are put on hard, so that the truck's wheelslock, and the truck skids to a stop. When the truck is empty, ithas a mass of 3000 kg, and it skids to a stop in 30 m. Whatdistance (in m) does the truck skid when it is loaded with a2,027-kg mass? (Ans: 3.00x10^1) 2) The terminal velocity of a 0.00008-kg drop is 1.5 m/s.Assuming a drag force F_D = -bv, what is the value of the constantb (in kg/s)? (Ans: 5.23x10^ -4) 3) The drag force on large objects moving through air is morenearly F_D = -bv^2. For this quadratic dependence on v, what is thevalue of the constant b (in Ns^2/m^2) if a 79-kg sky diver has aterminal velocity of 51 m/s? (Ans: 2.98*10^-1 )
Explanation / Answer
Initial speed u = 60 mph = 60 * 1609 m / 3600s = 26.82 m / s final velocity v = 0 distance S = 30 m empty mass m = 3000 kg from the relation v ^ 2- u ^ 2= 2aS Accleration a = [ v ^ 2- u ^ 2] / 2S = -11.98 m / s ^ 2 we know a = -g from this coeffciient of friction = a / g =1.223 After loading : -------------- total mass M = 3000+2027 = 5027 kg Acleration a ' = -g = -11.98 m / s ^ 2 reqiuired distance S ' = [ v ^ 2- u ^ 2 ] / 2a ' = 30 m (b). mass m = 0.00008 kg velocity v = 1.5 m / s At terminal velocity , weight = frictional drag from this frictional drag F= mg = 7.84* 10 ^ -4 N we know F = -bv from this b = F / v = 5.2266* 10 ^ -4 kg / s (3). mass m =79 kg velocity v = 51 m / s At terminal velocity , weight = frictional drag from this frictional drag F= mg = 774.2 N we know F = -bv ^2 from this b = F / v ^ 2 velocity v = 51 m / s At terminal velocity , weight = frictional drag from this frictional drag F= mg = 774.2 N we know F = -bv ^2 from this b = F / v ^ 2 = 0.2976Related Questions
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