# A conservative force of 10√2 N acting on a particle of mass m in XY – plane making an angle 45° ….

Q: A conservative force of 10√2 N acting on a particle of mass m in XY – plane making an angle 45° with X – axis as shown in the given figure. find work done by conservative force between origin and point A (1 m, 1 m , 0).

(a) Zero

(b) 10 J

(c) 20 J

(d) 30 J

Ans: (c)
Sol: $\displaystyle \vec{F} = 10\sqrt{2}cos45 \hat{i} + 10\sqrt{2}sin45 \hat{j}$

$\displaystyle \vec{F} = 10\sqrt{2}\frac{1}{\sqrt{2}} \hat{i} + 10\sqrt{2}\frac{1}{\sqrt{2}}\hat{j}$

$\displaystyle \vec{F} = 10 \hat{i} + 10 \hat{j}$

displacement vector , $\displaystyle \vec{s} = (1 \hat{i} + 1 \hat{j} + 0\hat{k}) – (0 \hat{i} + 0 \hat{j} + 0\hat{k})$

$\displaystyle \vec{s} = 1 \hat{i} + 1 \hat{j}$

$\displaystyle W = \vec{F}. \vec{s} = (10 \hat{i} + 10 \hat{j}).(1 \hat{i} + 1 \hat{j})$

W = 20 J