question_answer

A force \[\operatorname{F}=- K \left( y\hat{i} + x\hat{j} \right)\] (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 particle is

A) \[- 2K{{a}^{2}}\]

B) \[2K{{a}^{2}}\]

C) \[- K{{a}^{2}}\]

D) \[K{{a}^{2}}\]

Correct Answer: C

Solution :

[c] The expression of work done by the variable force F on the particle is given by \[\operatorname{W}=\int{\vec{F}.\overrightarrow{dl}}\] In going from (0,0) to (a, 0), the coordinate ofx varies from 0 to 'a', while that of .y remains zero. Hence, the work done along this path is: \[{{\operatorname{W}}_{1}}=\int_{0}^{a}{\left( -kx\hat{j} \right).dx\hat{i}=0}[\because \hat{j}.\hat{i}=0]\] In going from \[\text{(}a,0\text{) }to\text{ (}a,a\text{)}\] the coordinate of-c remains constant \[(=a)\] while that of y changes from 0 to 'a'. Hence, the work done along this path is \[{{\operatorname{W}}_{2}}=\int_{0}^{a}{[-K(y\hat{i}+a\hat{j}).dy\hat{j}]=ka}\int_{0}^{a}{dy=-k{{a}^{2}}}\] Hence, \[W={{W}_{1}}+{{W}_{2}}=-k{{a}^{2}}\]