Question

force acting on a particle is two ICAP + 3 J cap Newton works done by this force is zero when a particle is moved up along the line 3 Y + kx is equal to 5 then the value of k is

Answer 1

Force acting on a particle is $(2\hat i+3\hat j)$ Newton. Work done by this force is zero when a particle is moved up along the line $3y+kx=5$ then the value of k is 2

Explanation:

The force acting on the particle

$\vec F=(2\hat i+3\hat j)$ N

On moving this force up along the line $3y+kx=5$ is zero

This will be possible only when the line of action of the force is perpendicular to the line $3y+kx=5$

Since the position vector of the force is w.r.t. the origin

Therefore, the slope of the force vector will be

$m_1=\frac{3-0}{2-0}$

$\implies m_1=\frac{3}{2}$

The slope of the line $3y+kx=5$ is

$m_2=-\frac{k}{3}$

Since the line of action of force and the given line along which the force is moved are perpendicular

Therefore,

$m_1m_2=-1$

$\implies \frac{3}{2}\times(-\frac{k}{3})=-1$

$\implies \frac{k}{2}=1$

$\implies k=2$

Hope this is helpful.

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