Answer to Question #229930 in Chemical Engineering for Lokika

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+y\u00b2j + zk"


"x\u00b2+y\u00b2 = 4\\\\\nx\u00b2 = 4-y\u00b2\\\\\nx = \\sqrt{4-y\u00b2}"


"F = -(4-y\u00b2)^{1\/2} \u00d7yi + y\u00b2j +zk\\\\"

dw = F.dx

In j direction

"W=\u222b^2_0y(\\sqrt{4-y\u00b2})dy =8\/3\\ J"

(in z direction work done is always zero)


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