Initially, when the body of mass 2 Kg is moving on a rough surface with an acceleration of 5 ms-2. Then,
ma1=F−f
10=F−f.............(1)
When the horizontal force is doubled, it gets an acceleration of 18 ms-2. Then,
ma2=2F−f
36=2F−f........(2)
Solving equation (1) and (2)
36=2F−f
10=F−f
f=16N
∵f=μR
μ=fR
μ=fmg
μ=1620
μ=0.8