Answer to Question #259173 in Mechanics | Relativity for Farjana Akter

Question

Find the total work done in moving a particle in a force field given by F=3xyi-5zj+10xk along the curve x=t^2+1, y=2t^2, z=t^3 from t=1 to t=2.

Answer 1

Given,

$F=(3xyi-5zj+10xk)\\x=t^2+1\\y=2t^2\\z=t^3$

We have,

$\bar F\cdot \bar{dr}=(3xyi-5zj+10xk)\cdot(dxi+dyj+dzk)$

$= 3xydx-5zdy+10xdz\\=3(t^2+1)\cdot(2t^2)(2tdt)-5(t^3)(4tdt)+10(t^2+1)(3t^2dt)\\=(12t^5+12t^3-20t^4+30t^4+30t^2)dt\\=(12t^5+10t^4+12t^3+30t^2)dt$

So, Work Done = $\int_c \bar F\cdot \bar{dr}$ $=\int_1^2(12t^5+10t^4+12t^3+30t^2)dt\\$

$\\=[\frac{12t^6}{5}+\frac{10t^5}{5}+\frac{12t^4}{4}+\frac{30t^3}{3}]_1^2\\=[2t^6+2t^5+3t^4+10t^3]_1^2\\=[2(2)^6+2(2)^5+3(2)^4+10(2)^3]-[2+2+3+10]\\=303$