Jack and Jill are maneuvering a 3400 kg boat near a dock. Initially the boat\'s
ID: 2171453 • Letter: J
Question
Jack and Jill are maneuvering a 3400 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is 1.0 m/s. As the boat moves to position < 4, 0, 2 > m, Jack exerts a force of < -400, 0, 200 > N, and Jill exerts a force of < 160, 0, 320 > N.How much work does Jack do?
WJack = J
How much work does Jill do?
WJill = J
Which person exerted a force perpendicular to the displacement of the boat?
neither Jack nor Jill
Jill
both Jack and Jill
Jack
What is the final speed of the boat?
vf = m/s
Explanation / Answer
Good question. Let'me try help you. Unfortunately I am not sure, but I let some ideas. attempt 1. First, let's try discover the equation of the line (Consider travel straight). = + L Do you know vector calculus? Jack's Work = scalar F*s = -400*6 + -3*200 = -3000J Jill's Work = scalar F*s = 160*6 + 320*(-3) = 0 ========================= attempt 2. I'll try again (using just school): Distance between < 2, 0, 3 > and < 8, 0, 0 > = sqrt(6+0+9) = sqrt(15) = 3,873m Jack's force (module) = sqrt (400² + 200²)= 447,21 N But we need know the projection of forces in the direction of movement: F (Jack) = proj F, s = (vector F*s) / module(s) = . Jack's work = F*s = 0 F (Jill) = proj F, s = (vector F*s) / module(s) = . Jill's work = F*s = 357,77*3,873 = 1385,64J ======================== V0 = 1,8m/s to where? F=m*a a=F/m = v² = v0² + 2as v² = 3,24+ 2*0,12*3,873 => v = 2,04m/sRelated Questions
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