An automobile driver going at a speed of 24 km/h slams on the brakes and brings

Question

An automobile driver going at a speed of 24 km/h slams on the brakes and brings the car to a complete stop in 25 s. If the mass of the automobile is 1200 kg the magnitude of the average force exerted by the braking mechanism is: 6.5 times 10^2 N 7.2 times 10^2 N 5.4 times 10^2 N 3.2 times 10^2 N None of the above A 10 gram bullet traveling at 60 m/s hits and penetrates a thick wooden board to a depth of 5.0 cm. The average force exerted by live wooden board on the bullet is: 4.0 times 10^2 N 3.6 times 10^2 N 2.4 times 10^2 N 8.5 times 10^2 N None of the above A superhero sees an uncontrolled car traveling at a high speed of 180 km/h and, sensing the danger, stops the car within 10 m. If the car has a mass of 1600 kg the average force exerted by the superhero is: 125 kN 144 kN 160 kN 200 kN None of the above A particle slides down a 45 degree inclined plane that has a coefficient of kinetic friction coefficient of 0.48. The acceleration of the panicle down the incline is: 0.76 m/s^2 2.3 m/s^2 3.6 m/s^2 5.4 m/s^2 None of the above A boy and his bicycle have a total mass of 60.0 kg. At the top of the hill his speed is 5.0 m/s. His speed triples as he rides down the hill. Assume that there no losses of mechanical energy. What is the height of the top of the hill? 11.5 m 12.0 m 15.0 m 14.6m 10.2 m

Explanation / Answer

Given
1. d
   mass of automobile m = 1200 kg ,moving with speed v = 24 kmph = 24*5/18 = 6.67 m/s

time taken to come to rest after applying brakes is t = 25 s

we know that the change in momentum of the automobile = applied force (brake)

   m (dV/dT) = F

   F = 1200*(0-6.67)/25 N

   F = -320.16 N ------------->>>Answer
-ve sign indicates the brake force is acting in the -ve direction

2.b
given

mass of bullet m = 10 grams, speed v = 60 m/s, displacement s = 5 cm before come to rest

work done by the wooden board = change in k.e of bullet

       F*s = 0.5*m(v2^-v1^2)

       F = 1/s(0.5*m(v2^-v1^2))

       F = (1/0.05)(0.5*0.01(0^2-60^2)) N

       F = -360 N --------------->>> Answer

the force exerted by the wooden block is -360 N

3.d
mass of car m = 1600 kg, v= 180 kmph = 180*5/18 = 50 m/s , s = 10 m

   F*S = dk.e

   F = (0.5*m(v2^2- v1^2))/s

   F = (1/10)(0.5*1600(0^2-50^2))
   F = -200000 N

   F = 200 kN ------------------->>> answer

4.

the force equation is

   mgsin theta - muek mg cos theta = ma

   a = g( sin theta - muek cos theta)

   a = 9.8(sin45 - 0.48 cos45) m/s2

   a = 3.6 m/s2 -------------->>> Answer
answer is option c

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