Answer
0.58
Work Step by Step
For constant acceleration,
$v^{2}-u^{2}=2as$
$\implies a=\frac{v^{2}-u^{2}}{2s}=\frac{(1.60\,m/s)^{2}-0^{2}}{2\times0.90\,m}=1.42\,m/s^{2}$
According to Newton's second law, $F_{net}=ma$
$\implies F-f_{k}=ma$
$\implies 25\,N-f_{k}=3.5\,kg\times1.42\,m/s^{2}$
Or $f_{k}=25\,N-(3.5\,kg\times1.42\,m/s^{2})=20.03\,N$
Recall that $f_{k}=\mu_{k}N$.
Knowing that $f_{k}=20.03\,N$ and $N=mg=3.5\,kg\times9.80\,m/s^{2}=34.3\,N,$ we get
$\mu_{k}=\frac{f_{k}}{N}=\frac{20.03\,N}{34.3\,N}=0.58$
