A body is sliding down a rough inclined plane of angle of inclination - for which coefficient of friction varies with distance y as - y - Ky- where K is constant- Here y is the distance moved by the body down the plane- The net force on the body is zero at y - A- Find the value of constant K-

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Standard XIII

Physics

Maximum Push Using Friction

A body is sli...

Question

A body is sliding down a rough inclined plane of angle of inclination $θ$ for which coefficient of friction varies with distance $y$ as $μ (y) = K y$ , where $K$ is constant. Here $y$ is the distance moved by the body down the plane. The net force on the body is zero at $y = A$ . Find the value of constant $K$ .

\frac{t a n θ}{A}

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A c o t θ

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\frac{c o t θ}{A}

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A t a n θ

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Solution

The correct option is A $\frac{t a n θ}{A}$
The downward force = $m g s i n θ$
The upward force = $μ m g c o s θ$
Net force
$f (y) = m g s i n θ - μ m g c o s θ$
= $m g (s i n θ - K y c o s θ)$
at $y = A, f (y) = 0$
$0 = s i n θ - K A c o s θ$
or $K = \frac{t a n θ}{A}$

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A body is sliding down a rough inclined plane of angle of inclination θ for which coefficient of friction varies with distance y as μ(y)=Ky, where K is constant. Here y is the distance moved by the body down the plane. The net force on the body is zero at y=A. Find the value of constant K.

The correct option is A tanθAThe downward force = mgsinθ The upward force = μmgcosθ Net force f(y)=mgsinθ−μmgcosθ =mg(sinθ−Kycosθ) at y=A,f(y)=0 0=sinθ−KAcosθ or K=tanθA

The correct option is A $\frac{t a n θ}{A}$
The downward force = $m g s i n θ$
The upward force = $μ m g c o s θ$
Net force
$f (y) = m g s i n θ - μ m g c o s θ$
= $m g (s i n θ - K y c o s θ)$
at $y = A, f (y) = 0$
$0 = s i n θ - K A c o s θ$
or $K = \frac{t a n θ}{A}$