A particle is moved along a path `AB - BC - CD - DE - EF - FA`, as shown in figure, in presence of a force `vecF = (alpha y hati + 2 alpha xhatj)N`, where x and y are in meter and `alpha = -1Nm^(-1)`. The work done on the particle by this by this force will be Joule.

Answer

0.75

Explanation

`vecF = (alpha Y hati + 2 alpha x hatj) N`
`dw = vecF cdot vecd r = (alpha hati + 2 alpha x hatj) cdot (d x hati + dy hatj) = alpha y dx + 2 alpha xdy`
For `AB : dy = 0, y = 1 " ":. " " W_(AB) = int_0^1 alpha dx = alpha`
For `BC : dx = 0, x = 1 " ":. " "W_(BC) = int_(1)^(0.5) 2 alpha dy = 2alpha (-0.5) = -alpha`
For `CD : dy = 0, y = 0.5 " ":. " " W_(CD) = int_(1)^(0.5) (0.5)alpha dx = (0.5 alpha)(-0.5) = 0.25 alpha`
for `DE : dx = 0, x = 0.5 " " :. " "W_(DE) = int_(0.5)^(0) 2alpha(0.5) dy = alpha(-0.5) = -0.5 alpha`
For `EF : dy = 0, y = 0 " ":. " " W_(FE) = 0`
For `FA : dx = 0, x = 0 " " :. " "W_(FA) = 0`
Total work = `alpha + (-alpha) + (-0.25 alpha) + (-0.5 alpha) + 0 + 0 = -0.75 alpha = -0.75 (-1) = 0.75`.