Question #229930

Find the work done by a force F= -xyi+y²j + zk in moving a particle over the circular path x² + y² = 4, z = 0 from (2,0,0) to (0,2,0)


1
Expert's answer
2021-08-28T06:16:32-0400

F=xyi+y2j+zkF= -xyi+y²j + zk


x2+y2=4x2=4y2x=4y2x²+y² = 4\\ x² = 4-y²\\ x = \sqrt{4-y²}


F=(4y2)1/2×yi+y2j+zkF = -(4-y²)^{1/2} ×yi + y²j +zk\\

dw = F.dx

In j direction

W=02y(4y2)dy=8/3 JW=∫^2_0y(\sqrt{4-y²})dy =8/3\ J

(in z direction work done is always zero)


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS