Ch. 5 Solutions

Ch. 5 Solutions - CHAPTER 5 PROBLEM SOLUTIONS Problem 4. A...

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CHAPTER 5 PROBLEM 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 a av = (0 ! v 0 ) = t , and the average net force on the passenger, while coming to rest, is F av = ma av = ! m v 0 = 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/s 2 when a 3.3-N force is applied. Object B accelerates at 2.7 m/s 2 when 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 N is the only force acting. (a) When applied to either object, Newton’s second law gives F = m A a A and F = m B a B , so m B = m A = a A = a B = (8.1 m/s 2 ) = (2.7 m/s 2 ) = 3. (For constant net force, mass is inversely proportional to acceleration.) (b) When F is applied to the combined object, F = ( m A + m B ) a . Since F = m A a A , and m B = 3 m A , one finds a = F = ( m A + m B ) = m A a A = 4 m A = 1 4 (8.1 m/s 2 ) = 2.03 m/s 2 . (Note: It was not necessary to calculate the masses, which are m A = (3.3 N) = 2 ) = 0.407 kg , and m B = (3.3 N) = (2.7 m/s 2 ) = 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, F av » a av . To stop a car in a given distance, ( x ! x 0 ), a av = (0 ! v 0 2 ) = 2( x ! x 0 ), so F av » v 0 2 . Doubling v 0 quadruples the magnitude of F av , 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 " 10 3 N = 1400 kg = ! 5 . 71 m / s 2 . (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 v 0 x = 40 km/h can stop in a minimum distance found from Equation 2-11 and the maximum deceleration just calculated: x ! x 0 = ! v 0 x 2 = 2 a = ! (40 m = 3.6 s) 2 = 2( ! 5.71 m/s 2 ) = 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.
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