CHAPTER 5PROBLEM SOLUTIONS Problem 4. A car leaves the road traveling at 110 km/h and hits a tree, coming to a complete stop in 0.14 s. What average force does a seatbelt exert on a 60-kg passenger during this collision? Solution Assume that the seatbelt holds the passenger firmly to the seat, so that the passenger also stops in 0.14 s without incurring any secondary impact. Then the passenger’s average acceleration is aav=(0!v0)=t, and the average net force on the passenger, while coming to rest, is Fav=maav=!mv0=t=!(60 kg)(110=3.6)(m/s)=(0.14 s)=!13.1 kN, or about 1.5 tons. (Here, we used the one-dimensional form of Newton’s second law. The minus sign indicates that the force is opposite to the direction of the initial velocity. It is reasonable to assume that this is a component of the force exerted by the seatbelt.) Problem 7. Object A accelerates at 8.1 m/s2when a 3.3-N force is applied. Object B accelerates at 2.7 m/s2when the same force is applied. (a) How do the masses of the two objects compare? (b) If A and B were stuck together and accelerated by the 3.3-N force, what would be the acceleration of the composite object? Solution In this idealized one-dimensional situation, the applied force of F=3.3 Nis the only force acting. (a) When applied to either object, Newton’s second law gives F=mAaAand F=mBaB, so mB=mA=aA=aB=(8.1 m/s2)=(2.7 m/s2)=3.(For constant net force, mass is inversely proportional to acceleration.) (b) When Fis applied to the combined object, F=(mA+mB)a.Since F=mAaA, and mB=3mA,one finds a=F=(mA+mB)=mAaA=4mA=14(8.1 m/s2)=2.03 m/s2. (Note: It was not necessary to calculate the masses, which are mA=(3.3 N)=(8.1 m/s2)=0.407 kg, and mB=(3.3 N)=(2.7 m/s2)=1.22 kg.)Problem 9. By how much does the force required to stop a car increase if the initial speed is doubled and the stopping distance remains the same? Solution The average net force on a car of given mass is proportional to the average acceleration, Fav» aav.To stop a car in a given distance, (x!x0), aav=(0!v02)=2(x!x0),so Fav» v02.Doubling v0quadruples the magnitude of Fav, a fact that is important to remember when driving at high speeds. Problem 11. The maximum braking force of a 1400-kg car is about 8.0 kN. Estimate the stopping distance when the car is traveling (a) 40 km/h; (b) 60 km/h; (c) 80 km/h; (d) 55 mi/h. Solution The maximum braking acceleration is a=F=m=!8.0"103N=1400 kg=!5.71 m/s2.(We expressed the braking force as negative because it is opposite to the direction of motion.) (a) On a straight horizontal road, a car traveling at a velocity of v0x=40 km/hcan stop in a minimum distance found from Equation 2-11 and the maximum deceleration just calculated: x!x0=!v0x2=2a=!(40 m=3.6 s)2=2(!5.71 m/s2)=10.8 m.For the other initial velocities, the stopping distance is (b) 24.3 m, (c) 43.2 m, and (d) 52.9 m.
has intentionally blurred sections.
Sign up to view the full version.