Question

a wooden block of mass 0.6kg is on a rough horizontal surface. a force of 12N is applied to the block and it accelerates at 4m/sec^2 .what is the magnitude of the frictional force acting on the block

Answer 1

Given:

A wooden block of mass 0.6kg is on a rough horizontal surface. a force of 12N is applied to the block and it accelerates at 4m/s².

To find:

Magnitude of frictional force ?

Diagram:

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(15,15){\sf{m}}}\put(-20,-0.1){\line(1,0){55}}\multiput(-19,0)(3,0){19}{\qbezier(0,0)(-1,-1)(-2,-2)}\put(15,7.5){\vector(1,0){15}}\put(7.5,0){\vector(0,-1){15}}\put(4,-20){\sf{mg}}\put(7.5,15){\vector(0,1){15}}\put(6.3,31.5){\sf{N}}\put(0,0.2){\vector(-1,0){15}}\put(-17,1.3){\sf{f}}\put(32,6){\sf F}\end{picture}$

Calculation:

Let the coefficient of kinetic friction be $\mu:$

According to the FBD of the block:

$\rm \therefore \: F - f = ma$

$\rm \implies\: 12 - f = (0.6) \times 4$

$\rm \implies\: 12 - f = 2.4$

$\rm \implies\: f = 12 - 2.4$

$\rm \implies\: f = 9.6 \: N$

So, the frictional force acting on the block is 9.6 N.