A force F is related to the position of a particle by the relation $F = (10 x^{2}) N$ . The work done by the force when the particle moves from x = 2 m to x = 4 m is :

\frac{56}{3} J

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560 J

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\frac{560}{3} J

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\frac{3}{560} J

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Solution

The correct option is C $\frac{560}{3} J$
As we know that,

Given F = $10 x^{2}$

And x =2 to x =4.

Work done is based on the integration of force with distance as follows:

W= $\int_{2}^{4} F d x$

= $\int_{2}^{4} 10 x^{2} d x$

= $[\frac{(10 x^{3})}{3}]_{2}^{4}$

= $\frac{10}{3} [4^{3} - 2^{3})]$

= $\frac{10}{3} \times 56$

= $\frac{560}{3} J$

Thus option C is correct.

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Similar questions

F = k x^{2}

x = 0 m

x = 4 m

r_{1}

r_{2}

x^{2}

\to F = (- 6 x^{3}^i)

x = 4

x = - 2

\to F = (- 6 x^{3}^i) N

x = 4 m

x = - 2 m

¯ F = (- 6 x^{3} i) N

x = a m t o x = - 2 m

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