Question 56: Laws of Motion

Question

An inclined plane is bent in such a way that the vertical cross-section is given by $$y = {{{x^2}} \over 4}$$ where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction $$\mu$$ = 0.5, the maximum height in cm at which a stationary block will not slip downward is _________ cm.

Accepted Answer

The graph for given equation is shown below
At maximum height, the slope of tangent drawn,
$$	an 	heta  = {{dy} \over {dx}} = {{2x} \over 4} = {x \over 2}$$ [$$\because$$ $$y = {{{x^2}} \over 4}$$]
$$ \Rightarrow 0.5 = {x \over 2}$$ ($$\because$$ $$\mu$$ = tan$$	heta$$)
$$\Rightarrow$$ x = 1 m
$$	herefore$$ $$y = {{{x^2}} \over 4} = {1 \over 4}$$ = 0.25 m = 25 cm

Laws of Motion

Integer / Numeric Answer Type