A block mass `m = 2 kg` is accelerating by a force `F = 20 N` applied on a smooth light pully as shown in fig 7.86 If the coefficient of kinetic friction between the block and the surface is `mu = 0.3` find its acceleration

Explanation

From the FBD of pulley shown in fig (a) .

`F - 2T = m_(p)a_(p) :. T = (F)/(2)`….(i)(because `m_(p) = 0)`
From FBD of the block a fig (b)
`N = mg` …(2)
`:. T = f_(k) = ma`
i.e.`T - mu_(k) N = ma` ...(iii)
From (i) ,(ii) and (iii)
`(F)/(2)mu_(k) mg = ma :. a = ((F)/(2) - mu_(k) mg)/(m) = (F)/(2m) - mu kg `
`a = ((20)/(2 xx 2 - 3 xx 10)) ms^(-2) = ( 5- 3) ms^(-2) = 2 ms^(-2)`