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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Application +4 | -1
The force on a particle as the function of displacement x{x} (in x/{x} / direction) is given by
F=10+0.5x{F}=10+0.5 {x}

The work done corresponding to displacement of particle from x=0{x}=0 to x=2{x}=2 unit is
View answer and explanation
Correct option: C
21 J21 {~J}
Option C (21 J) is correct because it accurately calculates the work done by integrating the given force function over the specified displacement.

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