A block is placed on a ramp of parabolic shape given by the equation y=x^(2)//20. If mu_(s)=0.5, then the maximum height above the ground at which the block can be placed without slipping is

A block is placed on a ramp of parabolic shape given by the equation y

1.	A block is placed on a ramp of parabolic shape given by the equation y=x^(2)//20. If mu_(s)=0.5, then the maximum height above the ground at which the block can be placed without slipping is
Answer» Solution :Let the block can be placed on the ramp at a height h above the ground and `theta` is INCLINATION of the ramp at the position. In the position the component of weight along the slope of ramp is mg SIN `theta` downwards. The limiting frictional force is `= mu_(s)N=mu_(s)mg cos theta` In equilibrium, `mg sin theta = m_(s) mg cos theta qm sin theta = TAN theta = m_(s)=0.5` but, `y=X^(2)//20` Slope `tan theta = (dy)/(dx)=(2x)/(20)=(x)/(10)` `(x)/(10)=0.5 rArr x = 5` From the figure maximum height, `h=y_(MAX)=(x^(2))/(20)=(25)/(20)=1.25m`