Question

I Review Learning Goal: To use the principle of linear impulse to determine the change in velocity of a box subject to multiple forces Part A What is the magnitude of the normal force between the box and the ground? Express your answer to three significant figures with appropriate units. The block shown in (Figure 1) has mass m = 40 kg and is subjected to forces Fi = 300 N at an angle 0 = 20 ° below the horizontal and F = 60 N. The coefficient of kinetic friction between the box and the surface is x = 0.14. What is the velocity of the box after 13 s if it starts at rest? View Available Hint(s) TIMA = Value Units Figure < 1 of 1 Submit mg F: F2 Part B 11 UN What is the impulse in the horizontal direction due to Fı? Express your answer to three significant figures with appropriate units. View Available Hint(s)

Answer 1

Giver m=40kg fi=300N O=20 Fa=GON > Ficos. FR Uk=0.14 MIN t=13 sec (A) The of NP Magnitude Normal force my + Fisine 40x9-81) + (300) X Sin (20) N-4959 Ang 33 The impulse in the horizontal direction due to fi I=ficosoxt - 300 x Cos (20)x13 I: 3664.801 N-Sec Ans (6) The impulse due to fa I is (fa) xt GOX 13 >>I= 780 N-sec Ang (@) Impulse due to friction If = fxt = UK Nxt = 0:14 * (495) x13 If = 900.9 Ntsec sis

(f) velocity V=utat of Block offee t=13sec frorizontal xt - f Coso-fa- MIN xt 3000 Cos (20)-80-0.148295 X3 40 V= 3.815813 V=49.59789/3 Ans

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