A particle is displaced from a position vector 4i^+2j^m to 2i^+3k^m by applying a force of f→=3x2i^+2yj^N. The work done is

W=∫Fxdx+∫Fydy+∫FzdzW=∫42 3x2dx+∫20 2ydy+0W=3x3342+2y2220W=(8−64)+(0−4)=−56−4W=−60J

A particle is displaced from a position vector 4i^+2j^m to 2i^+3k^m by applying a force of f→=3x2i^+2yj^N. The work done is

W=∫Fxdx+∫Fydy+∫FzdzW=∫42 3x2dx+∫20 2ydy+0W=3x3342+2y2220W=(8−64)+(0−4)=−56−4W=−60J

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A particle is displaced from a position vector $(4 \hat{i} + 2 \hat{j}) m$ to $(2 \hat{i} + 3 \hat{k}) m$ by applying a force of $\vec{f} = (3 x^{2} \hat{i} + 2 y \hat{j}) N$ . The work done is

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-60J

60J

-120J

90J

answer is A.

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Detailed Solution

$\begin{array}{l} W = \int F_{x} d x + \int F_{y} d y + \int F_{z} d z \\ W = \int_{4}^{2} 3 x^{2} d x + \int_{2}^{0} 2 y d y + 0 \\ W = 3 {[\frac{x^{3}}{3}]}_{4}^{2} + 2 {[\frac{y^{2}}{2}]}_{2}^{0} \\ W = (8 - 64) + (0 - 4) = - 56 - 4 \\ W = - 60 J \end{array}$