A particle of mass m is displaced from a position P1 to P2 with position vectors ….

Problem 1:  A particle of mass m is displaced from a position P1 to P2 with position vectors $ \displaystyle \vec{r_1} = a\hat{i} + b\hat{j} + c\hat{k} $ and $ \displaystyle \vec{r_2} = c\hat{i} + a\hat{j} + b\hat{k} $ by a force $ \displaystyle \vec{F} = b\hat{i} + c\hat{j} + a\hat{k} $ . Find the work done by the force.

Solution:   If the displacement of the particle is $ \displaystyle \vec{S} $ , the work done W by the given force  is equal to  $ \displaystyle W = \vec{F}.\vec{S} $  , when  F is a constant force.

$ \displaystyle W = \vec{F}.\vec{S} $                …(1)

Where , $ \displaystyle \vec{F} = b\hat{i} + c\hat{j} + a\hat{k} $

and , $ \displaystyle \vec{S} = \vec{r_2} – \vec{r_1} $

$ \displaystyle \vec{S} = (c\hat{i} + a\hat{j} + b\hat{k}) – (a\hat{i} + b\hat{j} + c\hat{k}) $

$ \displaystyle \vec{S} = (c-a)\hat{i} + (a-b)\hat{j} + (b-c)\hat{k} $

$ \displaystyle W = (b\hat{i} + c\hat{j} + a\hat{k}).((c-a)\hat{i} + (a-b)\hat{j} + (b-c)\hat{k}) $

W = (c-a)b + (a-b)c + (b-c)a

W = bc – ab + ac – bc + ab – ac = 0

Net work done by the force for the given displacement is zero.

Next Page →