99 Chapter 3/4 Problem Solutions 3.36 We use the following notation: BS=vur velocity of boat relative to the shore BW=vur velocity of boat relative to the water, and WS=vur velocity of water relative to the shore. If we take downstream as the positive direction, then WS1.5 m s= +vur for both parts of the trip. Also, BW10 m s= +vur while going downstream and BW10 m s= −vr for the upstream part of the trip. The velocity of the boat relative to the shore is given by BW WSBS= +v v vur ur ur While going downstream, BS10 m s 1.5 m sv= + and the time to go 300 m downstream is ( )300 m26 s10+1.5 m sdownt = = When going upstream, BS10 m s 1.5 m s 8.5 m sv=− + =− and the time required to move 300 m upstream is 300 m35 s8.5 m supt−= =− The time for the round trip is ( )26 35 s 61 sdown upt t t= + = + = 3.41 Choose the positive direction to be the direction of each car’s motion relative to Earth. The velocity of the faster car relative to the slower car is given by FSFE ES= +v v vur ur ur, where FE60.0 km h= +vur is the velocity of the faster car relative to Earth and ES SE40.0 km h= − = −v vr r is the velocity of Earth relative to the slower car. Thus, FS60.0 km h 40.0 km h 20.0 km h= + − =+vur and the time required for the faster car to move 100 m (0.100 km) closer to the slower car is 3FS0.100 km 3600 s5.00 10 h 18.0 s20.0 km h 1 hdtv− = = = × = 100 CHAPTER 4 3.42 BC=vur the velocity of the ball relative to the car CE=vur velocity of the car relative to Earth 10 m s= BE=vur the velocity of the ball relative to Earth These velocities are related by the equation BE CEBC= +v v vur ur ur as illustrated in the diagram. Considering the horizontal components, we see that BE CEcos60.0v v° = or CEBC10.0 m s 20.0 m scos60.0 cos60.0vv = = =° ° From the vertical components, the initial velocity of the ball relative to Earth is BE BCsin60.0 17.3 m sv v= ° = Using ()2 202y y yv v a y= + ∆, with 0yv= when the ball is at maximum height, we find ( )( )()( )2220BE2max017.3 m s015.3 m2 22 9.80 m syyvvya g−−∆ = = = =− as the maximum height the ball rises. 4.6 52 277.5 10 N5.0 10 m s1.5 10 kgFam−Σ ×= = = ××, and 0v v at= + gives 02 280 km h 0 0.278 m s1 min7.4 min5.0 10 m s 1 km h 60 sv vta− −− = = = × 4.11 (a) From the second law, the acceleration of the boat is 22 000 N 1 800 N0.200 m s1 000 kgFam−Σ= = = f = 1 800 NF = 2 000 N+y+xFurfur60.0°vBEvBCvCEurururThe Laws of Motion 101 (b) The distance moved is ( )( )22 201 10 0.200 m s 10.0 s 10.0 m2 2x v t at∆ = + = + = (c) The final velocity is ()()200 0.200 m s 10.0 s 2.00 m sv v at= + = + = 4.13 Starting with 00yv= and falling 30 m to the ground, the velocity of the ball just before it hits is ()( )2 21 02 0 2 9.80 m s 30 m 24 m sy yv v a y= − + ∆ = − + − − = − On the rebound, the ball has 0yv= after a displacement 20 my∆ = +. Its velocity as it left the ground must have been ()( )2 222 0 2 9.80 m s 20 m 20 m sy yv v a y= + − ∆ = + − − = + Thus, the average acceleration of the ball during the 2.0-ms contact with the ground was ()4 22 1av320 m s 24 m s2.2 10 m s2.0 10 sv vat−+ − −−= = = + ×∆ × The resultant force acting on the ball during this time interval must have been ()()4 2 40.50 kg 2.2 10 m s 1.1 10 NF ma= = + × = + × or 41.1 10 N upward= ×Fr 4.17 From 0xFΣ =, 1 2cos30.0 cos60.0 0T T° − ° = or ()2 11.73T T= (1) Then 0yFΣ = becomes ()1 1sin 30.0 1.73 sin60.0 150 N 0T T° + ° − = which gives 175.0 N in the right side cableT = Finally, Equation (1) above gives 2130 N in the left side cableT = 150 NT2xy60.0°T130.0°urur102 CHAPTER 4 4.68 In the vertical direction, we have cos 4.0 0yF T mgΣ = ° − =, or cos 4.0mgT =° In the horizontal direction, the second law becomes: sin 4.0xF T maΣ = ° =, so 2sin 4.0tan 4.0 0.69 m sTa gm°= = ° =
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