Question

< Homework for MO2.5 Principle of Linear Impulse and Momentum 3 of 21 Review Learning Goal: To use the principle of linear impulse and momentum to relate a force on an object to the resulting velocity of the object at different times. The equation of motion for a particle of mass m can be written as Part A F = ma= m dv By rearranging the terms and integrating, this equation A jetliner of mass 8.54 104 kg is in level flight when it encounters a downdraft (a downward wind) that lasts for 1.50 s . The vertical component of the jetliner's velocity is 73.0 m/s after the downdraft subsides. What is the downdraft's average force, F, on the jetliner? Express your answer numerically in newtons to three significant figures. Figure < 1 of 1 View Available Hint(s) HF = 4.16x106 N F Submit Previous Answers Correct Correct answer is shown. Your answer 4.156-10% = 4.156X10 N was either rounded differently or used a different number of significant figures than required for this part.

Find part C

Answer 1

a llk = 0.205 Soletian's Data given F • 2700 N > F= fot N ولا د = 0.290 Sa 530° 0 50 - 30° > mass of crate (m) = 49 ng » Velocity finally = 2.43mis, then time (t) = ? free body diagram of block Yo N > As the block will move Kimetic friction in comes into picture + 50 f = lkn Quisbul 0800 but mg = f = lux mg caso > f = 0:205x49x9.81 X(03 30° f: 85.34 N & mg sino = 49x9.81 y sin 30 = 240.345 N → from Newton and law of motion F- f - Masino ma 5) 270- 85.34 - 240.345 = 49xa

a 2707 325.685 49 a = 5.51€ - 6.65 2 ) du = 5.51€ -6.65 at » Sar = $15.516-6.65.) elt = ) V=0 at t=0 j v = 2.43 mis / and at t=t 2.43 t » So sar $15.517-6.65) dy 0 0 9 2.43 = 5:51+2 6-65t 2 - 2.755+2_6.65t - 2.43 = 0 to = = 0.322 sec not possible tr=2.736 gec hence It 2.79 sec time taken to reach V=2.43mis