A body is moving down a long inclined plane of angle of inclination θ. The coefficient of friction between the body and the plane varies as μ$$=0.5x$$, where x is the distance moved down the plane. The body will have the maximum velocity when it has travelled a distance x given by $$x=N$$ $$tan$$⁡ θ . The value of N is

Question

A body is moving down a long inclined plane of angle of inclination θ. The coefficient of friction between the body and the plane varies as μ$$=0.5x$$, where x is the distance moved down the plane. The body will have the maximum velocity when it has travelled a distance x given by $$x=N$$ $$tan$$⁡ θ . The value of N is

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The acceleration of the body down the plane is gsin⁡ θ−μgcos ⁡θ$$=g($$$$sin$$ ⁡θ−μ$$cos$$⁡ θ$$)=g($$$$sin$$ ⁡θ−$$0.5$$ x $$cos$$θ$$)$$. Therefore, the body will first accelerate up to x <$$2$$ $$tan$$ θ. The velocity will be maximum at x $$=2$$ $$tan$$ θ, because for x > $$2$$ $$tan$$ θ, the body starts decelerating.

Friction

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The acceleration of the body down the plane is $gsin θ - μgcos θ = g (\sin θ - μcos θ) = g (\sin θ - 0.5 x cosθ)$ . Therefore, the body will first accelerate up to x <2 tan $θ$ . The velocity will be maximum at x =2 tan $θ$ , because for x > 2 tan $θ$ , the body starts decelerating.

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Friction

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The acceleration of the body down the plane is gsin⁡ θ−μgcos ⁡θ=g(sin ⁡θ−μcos⁡ θ)=g(sin ⁡θ−0.5 x cosθ). Therefore, the body will first accelerate up to x <2 tan θ. The velocity will be maximum at x =2 tan θ, because for x > 2 tan θ, the body starts decelerating.

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The acceleration of the body down the plane is $gsin θ - μgcos θ = g (\sin θ - μcos θ) = g (\sin θ - 0.5 x cosθ)$ . Therefore, the body will first accelerate up to x <2 tan $θ$ . The velocity will be maximum at x =2 tan $θ$ , because for x > 2 tan $θ$ , the body starts decelerating.