CGPET 2019 Solved Paper

Given, µ=

1

2

,m1=2kg,m2=4kg
F1=2N,F2=20N
The maximum force of friction that can occur between the two blocks, is
fmax=µmg ........(i)
Substitute given value in Eq. (i)
∴fmax=

1

2

×2×10=10N
and acceleration of the block system,
a=

total force

total mass

=

20+(−2)

4+2

=

+18

6

=3 m∕s2
The system of block has an acceleration in the left side. The force of friction opposing the motion of the 2kg block in the left side is
ma+ force acting in the direction of friction
=[(2×3)+2N]=8N
Thus, the force of friction acting between the two blocks is 8N.