Question

Find the work done by the force ​F = xyi +(y-x)j over the straight line from​ (-1,1)to (3,-3). The amount of work done is ___?

Answer 1

Answer:

amount of work done is$W = \frac{-4}{3}$      

Explanation:

Formula for work done by force field

$W = \int F. dr = \int_{t_o}^{t_1} f(r(t)) r'(t) dt$

where

r(t) is parametrization of line

as it is straight line so

$r(t) = 1- t) r_o + tr_1 0 \leq t \leq 1$

thus,

r(t) = (1-t)(-1,1) + t(3,-3)

    = (-1+t,1-t) + (3t - 3t)

   = (-1+t +3t, 1-t-3t)

r(t) = (4t -1, 1- 4t)

r'(t) = (4,-4)

putting value in above integral

$\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \int_{0}^{1} (-(4t -1)^2 , 2-8t).(4,-4) dt$

                                                  $= \int_{0}^{1} (-16 t^2 + 8t -1,2-8t) .(4,-4) dt$

                                                  $=\int_{0}^{1} (4(-16t^2 +8t -1) -4(2-8t)) dt$

                                                   $= 4[ -16 \frac{t^3}{3} + 16\frac{t^2}{2} - 3t]_{0}^{1}$

$\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \frac{-4}{3}$