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ID: 1957917 • Letter: #
Question
<p><img src="https://s3.amazonaws.com/answer-board-image/18c7b442-040c-48e6-b39f-5bf5f24b5cb3.jpg" alt="image from custom entry tool" /></p><p>If <img src="https://s3.amazonaws.com/answer-board-image/cramster-equation-20119235495863452353798791029763.gif" alt="" align="absMiddle" />, evaluate the work done by this force acting on a particle following a triangular path with vertices at (3,0,0), (6,0,0), and (3,4,0), where the positions<br />are given in meters.</p>
Explanation / Answer
F = (2x+y6)ˆi +(4y3x5)ˆj N
Workdone = F.ds (dot product of vectors F and ds)
Along x-axis from (3,0,0) to (6,0,0):
F may be written as F = 2x^i-(3x+5)^j and ds = dx^i
Workdone = F.ds = 2x dx = x^2 from 3 to 6 = 6^2-3^2 = 27 Joules
Along line 3y+4x=24
2x+y-6 = 2x+[(24-4x)/3]-6 = (6-2x)/3
4y-3x-5 = 4y-3[(24-3y)/4]-5 = (25y-92)/4
ds = dx^i+dy^j
workdone = F.ds = (6-2x)/3 dx + (25y-92)/4 dy , x varying from 6 to 3 and y from 0 to 4
= 3-42 = -39 Joules
Along line x=3
F = y^i + (4y-14)^j , ds = dx^i+dy^j
work done = F.ds = (4y-14)dy y varying from 4 to 0 = 40 joules
Total workdone = 27-39+40 = 28 Joules
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