compute work done by F = (2x-ysin(xy))i + (2-xsin(xy)j on a particle going from

Question

compute work done by F = (2x-ysin(xy))i + (2-xsin(xy)j on a particle going from (1,pi) ..... Please provide step by step solution

compute work done by F = (2x-ysin(xy))i + (2-xsin(xy)j on a particle going from (1,pi) ..... Please provide step by step solution Compute the work done by F = (2x ? y sin xy)i + (2 ? x sin xy)j on a particle going from to (0, -1) along a path made of 3 straight lines: one from to (12, 0), one from (12, 0), one from (12, 0) to ( -11, 0) and one from (-11, 0) to (0, -1).

Explanation / Answer

dW = F . r'(t) dt

We first have to parameterize the curve. We have

r(t) = <1,pi> + [<0,-1> - <1,pi>]t = <1-t,pi-(1+pi)t>

r'(t) = -i -(1+pi)j

Taking the dot product, we get

F . r'(t) = -(2x-ysin(xy)-(1+pi)(2-xsin(xy))

x=1-t,y =pi -(1+pi)t

just put the value and integrate .

or u can do it as

work done = F.S (its a dot product of two vectors)

here F has been given
take S in three steps
S1 = 11 i - pi j
S2 = -23i + o j
S3 = 11i - 1 j
So total work done will be
W = F.S1 + F.S2 + F.S3

you can also take help from

http://www.ltcconline.net/greenl/courses/202/vectorIntegration/lineIntegrals.htm

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