A person of mass `75 kg` works in a where his job is to arrange heavy articles in a stone - room As shown in figure he place a packing of mass `20 kg` on a ramp that is inclined to the horizontal at `37^(@)` and pushed it up the ramp with an acceleration `1.5ms^(-2)` .This is the minimum acceleration for the packege to reach the top to the ramp.THe coefficient of static and friction between the shoes of the person and the ground is `0.8` and coefficient of kinetic friction for the motion of package on the is `0.5`

Force applied by the person on the package is

Options

A. `120 N`
B. `250 N`
C. `210 N`
D. None of these

Answer

d

Explanation

Friction force acting on the block
`f_(k) = mu_(1) mg cos theta`
Let man applied force F
`F - f_(k) - mg sin theta = ma F = 230N`
Limitimg friction between man and ground
`f_(k) = mu_(2)(Mg + F sin theta) = 0.8 (750 + 230 xx 3//5) = 710.4 N`
But driving force acting on the man
`f_(x) = F cos theta 184 N`
Here `f_(s) gt F_(x)`
Hence man will not slide on ground and friction force will be static nature Hence , frcition force acting on the man is `f = 184 N`
For sliding the block on plane offered by man should be same as before as acceleration is not changing Hence `F = 230 N`
If friction coefficient changed to `mu = 0.2` the limiting friction will reduce and changed to
`f_(s) = 0.2 (Mg + F sin theta) = 177.6 N`
But driving force acting on man in `F cos theta = 184 N` block is greater than maximum value of tyhe friction force that be developed between ground and man Hence man and will slide and friction force will be equal to `f = f_(lim) = 177.6 N`