hw6 - PROBLEM_13.11‘9 v A\2—kg particle is acted upon...

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Unformatted text preview: PROBLEM_13.11‘9 _- v ; A\2—kg particle is acted upon by a force F = ~2t2i + (3 —- t) j, where F is expressed in newtons and t in seconds. Knowing that the velocity of the particle is v; = (3 m/s)i at t = 0, determine (a) the time at which the velocity is parallel to the y axis, (b) the corresponding velocity of the . particle. , SOLUTION (a) ‘ mvo + Eth = m(vxi + vyj) But vx = 0, if velocity is parallel to y-axis V (kag~)3i + flare +(3 .—. t)j] dt =él2Vyj , 2t3 2 ' 6i'— ( 3‘)i+[3t —-t2—]j =‘ZVyj 2t3 . t2 . . [6 — 3-} + [3t — 3].} = 213),] Since the x component of the velocity is zero 3 6—ZL=0 23:9 3 ,. (b) Substitute r = 2.08 111(1) 0i + [3(208) — 4.0773 = 2vyj vy = 2.04 m/s 4 PROBLEM‘13.126 A ’440-kg sailboat with its occupants is running downwind at 12 km/h when its spinnaker is raised to increase its speed. Determine the net force provided by the spinnaker oVer the 107s interval that it takes for the boat to reaéh a speed of 18'km/h. ' * ‘ SOLUTION v1=12km/h = 3.33 m/s t1_2 = 10s v2 =18km/h = 5.00 m/s t:::7+t::-:~r7=+~:::7 WV, . aim mt; mv1 + impulse1_2 = mv2 m(333 m/s) + FN (10 s) = m(5.00 m/s) 10s N = 73.33 N FN = 73.3N4 Note: F N is the netforce provided by the sails. The force on the sails is actually greater and includes the force needed to overcome the water resistance of the hull. PROBLEM 13.140 The last segment of the triple jump track-and-field event is the jump, in 9.14m 35° which the athlete makes a final leap, landing in a sand-filled pit. Assuming that the velocity of an 84-kg athlete just before landing is 9.14 m/s at an angle of 35° with the horizontal and that the athlete comes to a complete stop in 0.22 s after landing, determine the horizontal component of the average impulsive force exerted on his feet during landing. Landing pit SOLUTION (a) . l t Wu; am: 3‘ At = 0.22 mv1 + (P — W)At ; mvz Horizontal components T 84(9.14cos35°) — PH(0.22) = 0 PH = 2858.69 kg-m/s2 PROBLEM 13.1 57 A 6.00-g ball A is moving with avelocity VA when it is struck by a 1.2-kg ball B” which has a velocity VB of magnitude DB = 6 m/s. Knowing that v. the velocity of ballB is zero after impact and that the coefficient of B restitution is 0.8, determinetheavelocity of ball A (a) befozevimpact, (b) after impact. SOLUTION From conservation of momentum ‘ _ ‘r. 'I J; mAVA + vaB — mAVA + mBVB b.6151 —1.2(6)‘= —0.6v';1 + 0 '. VA , v' =O.8v +4.8 vA+6 A A From restitution 0.8 = (a) Velocity of A before impact from Equations (1) and (2) I 0.6vA — 742 = 40.6(0.8vA +4.8)”: -0.48vA — 2.88 1.08vA = 4.32 (b) Velocity of A after impact v'A = 0.8(4)*+ 4.8‘ ...
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