A 60 kg block is pushed up on inclined plane by means of a horizontal push P as shwon in the figure . The coefficients of friction between the incline and block are `mu_(s) = 0.6` and `mu_(k) = 0.4` and the ramp makes an angle of `30^(@)` with the horizontal .
(a) What value of P is required to move the block at a constant speed of `0.20` m/s along the incline ?
(b) If the person should stop for rest and let P = 0 , does the block slide back on the incline ?

Explanation

From the force diagrams as shown in figure :

x-components : `F_("fric") + W "sin" theta - P "cos" theta = 0 " " …….`(i)
y-components : `N - W "cos" theta - P "sin" theta = 0 " " ….. `(ii)
where `F_("fric") = mu_(k) N`
From Eqns. (i) and (ii) ,
`mu_(k) (W "cos" theta + P "sin" theta ) + W "sin" theta - P "cos" theta = 0`
or `P[mu_(k) "sin" theta - "cos" theta ] = -W [mu_(k) "cos" theta + "sin" theta ]`
Hence we obtain , `P = ("sin" theta + mu_(k) "cos"theta )/("cos" theta - mu_(k) "sin" theta)W = 747 . 24N`
If force P is removed , the block will have a tendency to slide down due to W sin `theta`(component of weight along incline ).
Maximum possible force of friction is `mu_(s) N`
(i) If `W "sin " theta gt mu_(s) N` , the block slips down .
(ii) If W sin `theta lt mu_(s)N` , the block will stay in equilibrium in this situation .
As here W sin `theta = W xx (1)/(2) lt mu_(s) N = 0.6 xx W "cos" theta= 0.6 W xx (sqrt3)/(2)`, [from (i) , putP = 0]
therefore , the block will stay in equilibrium .