Question

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

• A. $\frac{4}{9}$
• B. $\frac{9}{4}$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

$4$

Let ${\theta }_1$ is the angle of $\overrightarrow{F}$ with positive x-axis

$\therefore \tan {\theta }_1=\frac{F_y}{F_x}=\frac{15}{20}=\frac{3}{4}=m_1$ (say)

Slope of given line, $m_2=-\frac{\alpha }{3}$

$W=0$ if $\overrightarrow{F}\perp \overrightarrow{S}$

or $m_1{m}_2=-1$

$\therefore \left(\frac{3}{4}\right)(-\frac{\alpha }{3})=-1$

$\therefore \alpha =4$.