NEET UG PHYSICS MCQ: WORK-ENERGY-POWER

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A force

\mathrm{\vec F} = - K\left( {y\hat i + x\hat j} \right)

(where K is a positive constant) 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 y-axis to the (a,a). The total work done by the force

\mathrm{\vec F}

on the particle is

View answer and explanation

Correct option: C

- K{a^2}

Option C (

- K{a^2}

) is correct because the work done is path-dependent for this non-conservative force, and the integral along the specified path yields this result.

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\mathrm{\vec F} = \left( {\hat i + 2\hat j + 3\hat k} \right)

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{\mathrm{\vec r}_1} = \left( {\hat i + \hat j + \hat k} \right)

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