Kinematics Question 114
Question: A force $ \overrightarrow{F}=-K(y\hat{i}+x\hat{j}) $ (where K it’s a positive constant) acts on a particle moving in the x-y 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 forces $ \overrightarrow{F} $ on the particle is [IIT-JEE 1998]
Options:
A) $ -2Ka^{2} $
B) $ 2Ka^{2} $
C) $ -Ka^{2} $
D) $ Ka^{2} $
Correct Answer: C For motion of the particle from (0, 0) to (a, 0) $ \overrightarrow{F}=-K(0\hat{i}+a\hat{j}) $ Displacement $ \overrightarrow{r}=(a\hat{i}+0\hat{j})-(0\hat{i}+0\hat{j})=a\hat{i} $ So work done from (0, 0) to (a, 0) it’s given by $ W=\overrightarrow{F}.\overrightarrow{r} $ $ =-Ka\hat{j}.a\hat{i}=0 $ For motion (a, 0) to (a, a) $ \overrightarrow{F}=-K(a\hat{i}+a\hat{j}) $ and displacement $ \overrightarrow{r}=(a\hat{i}+a\hat{j})-(a\hat{i}+0\hat{j})=a\hat{j} $ So work done from (a, 0) to (a, a) $ W=\overrightarrow{F}.\overrightarrow{r} $ $ =-K(a\hat{i}+a\hat{j}).a\hat{j}=-Ka^{2} $ So total work done $ =-Ka^{2} $Show Answer
Answer:
Solution:
$ \Rightarrow \overrightarrow{F}=-Ka\hat{j} $
