AIIMS2017Physics-Work, Energy and Power

AIIMS 2017 Physics Work Done by a Variable Force MCQ Question

Type: MCQ-numerical-Medium-Class 11

A force F=k(yi^+xj^)\vec{F} = -k\left(y\hat{i} + x\hat{j}\right), where kk is a positive constant, acts on a particle moving in the xyxy-plane. Starting from the origin, the particle is taken along the positive xx-axis to the point (a,0)(a,0) and then parallel to the yy-axis to the point (a,a)(a,a). The total work done by the force on the particle is

A

2ka2-2ka^2

B

2ka22ka^2

C

ka2-ka^2

D

ka2ka^2

Correct Answer

Option A

Detailed Explanation

The expression for work done, W=0ak(ydx+xdy)W = \int_0^a -k (y \, dx + x \, dy), simplifies to W=k0a(xy)dW = -k \int_0^a (xy) \, d, which evaluates to k(a2)-k(a^2) when considering the limits of integration. This indicates that the work done by the force FF over the displacement drdr results in a negative value, reflecting energy loss in the system. The other options are not applicable as they do not provide relevant values or interpretations related to the work done in this context, thus reinforcing the importance of understanding vector calculus in physics.

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