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Vector Component

If a force F ...

Question

If a force $F$ acts along $X$ axis and $F$ varies as $F = 6 x^{2} + 10$ then work done by force to move block along $x^{2} + y^{2} = 25$ from (0, -5) to (0, +5)

125 J

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-125 J

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300 J

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0 J

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Solution

The correct option is D 0 J
Net displacement along the direction of force i.e. $x$ -axis is $0$ hence, $W = 0$ J.

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Similar questions

F

X

F

F = 6 x^{2} + 10

x^{2} + y^{2} = 25

F

X

F

F = 6 x^{2} + 10

x^{2} + y^{2} = 25

F = - 6 x^{3}

F = \frac{4}{x^{2}} (x \neq 0) .

F = 7 - 2 x + 3 x^{2}

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