A particle moves under the action of a force $F = 20 \hat{i} + 15 \hat{j}$ along a straight line $3 y + α x = 5$ , where, $α$ is a constant. If the work done by the force F is zero, then the value of $α$ is

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$\frac{4}{9}$

$\frac{9}{4}$

$3$

$4$

answer is D.

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Detailed Solution

$Let θ_{1} is the angle of F with positive x -axis$

$∴ \tan θ_{1} = \frac{F_{y}}{F_{x}} = \frac{15}{20} = \frac{3}{4} = m_{1} (say)$

$Slope of given line, m_{2} = - \frac{α}{3}$

$\begin{array}{l} W = 0 if F ⊥ S \\ or m_{1} m_{2} = - 1 \\ ∴ (\frac{3}{4}) (- \frac{α}{3}) = - 1 \\ ∴ α = 4 \end{array}$

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