The work done by a force $¯ F = (- 6 x^{3} i) N$ in displacing particle from $x = a m t o x = - 2 m$ is

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Solution

$W = \int_{_{1}}^{_{2}} \to F . \to d r$

here,
$W = \int_{a}^{- 2} F_{x} d x = - 6 \int_{a}^{- 2} x^{3} d x = \frac{- 6}{4} {[x^{4}]}_{a}^{4}$

$⟹ W = \frac{3}{2} (a^{4} - 16) J$

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Similar questions

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

x = 4 m

x = - 2 m

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

x = 4 m

x = - 2 m

(4 x^{2} + 3 x) N

x = 2 m

x = 3 m

(4 x^{2} + 3 x) N

x = 2 m

x = 3 m

Work done by a force, $F = (- 6 x^{3})^i N$ , in displacing a particle from x = 4m to x = –2m is:

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