A `2kg` block is in contact with a vertical wall having coefficient of friction `0.5` between the surfaces. A horizontal force of `40N` is applied on the block at right angles to the wall. Another force of `15N` is applied on the plane of the wall and at right angles to `40N` force. Find the acclecration of the block .

Explanation

Resultant of W = 20 N and 15 N
`F_(R)=sqrt(20^(2)+15^(2))=25N`
frictional force `f=muN=0.5xx40=20N`
This acts in a direction, opposite to 25N force.
This acts in a direction, opposite to 25N force.
`therefore` Net force acting on the block, `F_("net")=25-20=5`
`therefore` acceleration of the block `a=(5)/(2)=2.5ms^(-2)`