Work Energy And Power Question 233
Question: A force $ F=- K ( y\hat{i} + x\hat{j} ) $ (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
Options:
A) $ - 2Ka^{2} $
B) $ 2Ka^{2} $
C) $ - Ka^{2} $
D) $ Ka^{2} $
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Answer:
Correct Answer: C
Solution:
[c] The expression of work done by the variable force F on the particle is given by
$ 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: $ {W_1}=\int_0^{a}{( -kx\hat{j} ).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 $ {W_2}=\int_0^{a}{[-K(y\hat{i}+a\hat{j}).dy\hat{j}]=ka}\int_0^{a}{dy=-ka^{2}} $ Hence, $ W=W_1+W_2=-ka^{2} $
BETA
