CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
ktanθ
loader
C
cotθk
loader
D
kcotθ
loader

Solution

The correct option is A $$\dfrac{\tan\theta }{k}$$
The net downward force along the inclined plane on the body at a
distance $$x$$ 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
$$\sin\theta-kx_{0}\cos\theta=0$$          $$\therefore x_{0}=\dfrac{tan\, \theta }{k}$$
130046_7836_ans_9b5f1dd0447d4811ac6b0febc976430a.png

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image