Question

A block of mass $20$ kg is acted upon by a force $F=30N$ at an angle ${53}^0$ with the horizontal in downward direction as shown. The coefficient of friction between the block and the horizontal surface is $0.2$. The friction force acting on the block by the ground is $(g=10m/s^2)$

• A. 40.0 N
• B. 30.0 N
• C. 18.0 N
• D. 44.8 N

Answer 1

The correct option is C 18.0 N

Max. frictional force
$f_{max}=\mu N$
$=\mu (mg+Fsin{53}^0)$

$=0.2(20\times 10+30\frac{4}{5})=44.8N$

As $F\cos \theta =30\cos {53}^0=30\times \frac{3}{5}=18N<f_{max}$ friction force will also be $18$ N.