Answer to Question #205117 in Optics for anuj

Question

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.

Answer 1

Solution:-

we have the following data given

Force $F = xy\hat{i} + yz\hat{j} + xz\hat{k }$

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

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

Work done $W=\int F.dr$

$W=\int(xy\hat{i} + yz\hat{j} + xz\hat{k })(\hat{i} +4t\hat{j} + 3t^2\hat{k})dt$

$=\int xydt+yz\times 4t dt+xz\times 3t^2 dt$

Now, x = t, $y=t^2$ and $z=t^3$

$=\int t^3dt + 4t^6 dt+3t^6 dt$

$=\int 7t^6 dt + \int t^3 dt$

$=t^7+\frac{t^4}{4}$

Now, substituting the values of t = 1

$=\frac{5}{4} \ answer$