Question #205117

determine the work done by a force F = xyi + yzj + xzk in taking a particle along the path defined by the equation r(t)=ti + 2t²j + t³k, 0≤t≤1 from t=0 to t=2.


1
Expert's answer
2021-06-14T15:01:57-0400

Solution:-

we have the following data given

Force F=xyi^+yzj^+xzk^F = xy\hat{i} + yz\hat{j} + xz\hat{k }

Path equation r(t)=ti^+2t2j^+t3k^r(t) = t\hat{i} + 2t^2\hat{j} + t^3\hat{k}

d(r)=i^+4tj^+3t2k^dtd(r) = \hat{i}+4t\hat{j}+3t^2\hat{k}dt


Work done W=F.drW=\int F.dr

W=(xyi^+yzj^+xzk^)(i^+4tj^+3t2k^)dtW=\int(xy\hat{i} + yz\hat{j} + xz\hat{k })(\hat{i} +4t\hat{j} + 3t^2\hat{k})dt

=xydt+yz×4tdt+xz×3t2dt=\int xydt+yz\times 4t dt+xz\times 3t^2 dt

Now, x = t, y=t2y=t^2 and z=t3z=t^3

=t3dt+4t6dt+3t6dt=\int t^3dt + 4t^6 dt+3t^6 dt

=7t6dt+t3dt=\int 7t^6 dt + \int t^3 dt

=t7+t44=t^7+\frac{t^4}{4}

Now, substituting the values of t = 1

=54 answer=\frac{5}{4} \ answer

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