A block of mass 5 kg is kept on a rough horizontal surface. The coefficient of friction for the contact force block and surface are `mu_(s) = 0*1 and mu_(k) = 0*9 ` respectively . A horizontal force F given by F `0*5t` is applied at t = 0 . Here t represent time in seconds . Determine the force of friction and acceleration of the block at

(i) t = 1s (ii) t= 10 s (iii) t = 12 s

Explanation

The block will move when the external force F exceeds the force of limiting friction . Figure shows the FBD of the block .
As the block does not move along vertical ` N = " mg " = 50` newton
The force of limiting friction is ` mu _(s) N = 0*1 xx 50 = 5 N `
the block will move when F exceeds 5 N
(i) ` t = 1s , F = 0*5 t = 0*5 xx 1 = 0*5 N`
the block does not move
` :. F - f = 0 rArr f = 0*5 , a = 0 `
(ii) `t = 10 " s "F= 0*5 xx 10 = 5 N `
The block is in a state of limiting equilibrium but it still does no move
` rArr a = 0 , f = 5 N `
(iii) ` t = 12 s , F = 0*5 xx 12 = 6*0 N `
As F `gt` limiting force of friction , block will move in the direction of applied force.
` f_(k) = mu_(k) N = 0*09 xx 50 = 4*5 N`
` F - f _(k) = (6- 4*5)/5 = 0*3 "m//s^(2)`