Work Energy And Power Question 308
Question: A force $ \overset{\to }{\mathop{F}}=-k(y\hat{i}+x\hat{j}) $ 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 j’-axis to the point (a, a). The total work done by the force is
Options:
A) $ -2ka^{2} $
B) $ 2ka^{2} $
C) $ -ka^{2} $
D) $ ka^{2} $
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Answer:
Correct Answer: C
Solution:
[c] $ W_1\int_0^{a}{\overset{\to }{\mathop{F}}.\overset{\to }{\mathop{dx}}}=\int_0^{a}{-k(y\hat{i}-x\hat{j}).\hat{i}dx} $
$ =\int_0^{a}{-k(0\hat{i}}+x\hat{j}).\hat{i}dx=zero $
$ W\int_0^{a}{\overset{\to }{\mathop{F}}.\overset{\to }{\mathop{dy}}}=\int_0^{a}{-k(y\hat{i}}+x\hat{j}).\hat{j}dy $
$ =\int_0^{a}{-k(a\hat{i}}+a\hat{j}).\hat{j}dy $
$ =-ka\int_0^{a}{dy=-ka^{2}} $
Total work done, $ W=W_1+W_2=0-ka^{2}=-ka^{2} $
BETA
