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

Explanation

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 qmg 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`