Question

# A body is sliding down a rough inclined plane of inclination $$\theta$$, for which the coefficient of friction varies with distance $$x$$ as $$\mu (x)=kx$$, where $$k$$ is a constant. Here $$x$$ is the distance moved by body down the plane. The net force on the body down the plane will be zero at a distance $$x_{0}$$ given by

A
tanθk
B
ktanθ
C
cotθk
D
kcotθ

Solution

## The correct option is A $$\dfrac{\tan\theta }{k}$$The net downward force along the inclined plane on the body at adistance $$x$$ is:$$f(x)=mg\sin\theta-\mu mg\cos\theta$$        $$=mg(\sin\theta-\mu\cos\theta)$$        $$=mg(\sin\theta-kx\cos\theta)$$$$\therefore f(x)=0$$ at a value of $$x=x_{0}$$ given by$$\sin\theta-kx_{0}\cos\theta=0$$          $$\therefore x_{0}=\dfrac{tan\, \theta }{k}$$Physics

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