Question

# A force F=−K(y^i+x^j), where K is a positive constant, acts on a particle moving in the x - y plane. Starting from the origin, the particle is taken first 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
2Ka2
B
2Ka2
C
Ka2
D
Ka2

Solution

## The correct option is C −Ka2In going from (0,0) to (a,0), the x-coordinate varies from 0 to a while the y-coordinate remains zero. ∴ Work done by force F along this path is (∵y=0) W1=∫a0F.dx=∫a0−(Kx^j).dx^iW1=0 (∵^j.^i=0) In going from (a,0) to (a,a), the x-coordinate remains constant at x=a while y-coordinate changes from 0 to a. ∴ Work done by force F along this path is (∵x=a) W2=∫a0F.dy=∫a0−K(y^i+a^j).dy^j =−Ka∫a0dyW2=−Ka2 (∵^i.^j=0,^j.^j=1) The total work done is: W=W1+W2=0−Ka2=−Ka2.

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