Work Energy And Power Question 65
Question: A force $ F=-K(yi+xj) $ (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is [IIT 1998]
Options:
A) $ -2Ka^{2} $
B) $ 2Ka^{2} $
C) $ -Ka^{2} $
D) $ Ka^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
While moving from (0,0) to (a,0)
Along positive x-axis, y = 0
$ \therefore \ \vec{F}=-kx\hat{j} $ i.e. force is in negative y-direction while displacement is in positive x-direction. \ $ W_1=0 $
Because force is perpendicular to displacement
Then particle moves from $ (a,0) $ to $ (a,a) $ along a line parallel to y-axis $ (x=+a) $ during this $ \vec{F}=-k(y\hat{i}+a\hat{J}) $
The first component of force, $ -ky\hat{i} $ will not contribute any work because this component is along negative x-direction $ (-\hat{i}) $ while displacement is in positive y-direction (a,0) to (a,a).
The second component of force i.e. $ -ka\hat{j} $ will perform negative work \ $ W_2=(-ka\hat{j})\ (a\hat{j}) $ = $ (-ka)\ (a)\ =-ka^{2} $
So net work done on the particle $ W=W_1+W_2 $ = $ 0+(-ka^{2})=-ka^{2} $
BETA
