c
Let \(µ_s\) and \(µ_k\) be the coefficients of static and kinetic friction between the box and the plank respectively.
When the angle of inclination \(0\) reaches \(30°\), the block just slides, \(\mu_s=\tan\theta =\tan 30^o = \frac{1}{\sqrt3}=0.6\)

If \(a\) is the acceleration produced in the block, then
\(ma = mg\sin\theta -f_k\)
(where \(f_k \) is force of kinetic friction)
\(= mg\sin0 -μ_k N\)                   \((as\ \ f_k. = μ_k N)\)
\(= mg\sin0 - μ_kmg\cos\theta\)            \((as\ \  N = mgcos\theta)\)
\(a = g(\sin\theta - μ_k\cos\theta)\)
As \( g = 10 ms^{- 2}\) and \(0 = 30°\)
\(a = ( 10 ms^{- 2})(sin30^o -μ_kcos30^o)\)  ..........\((i)\)

If \(s\) is the distance travelled by the block in time \(t\), then

\(S=\frac{1}{2} at^2\)                \((u=0)\)

\(a=\frac{2s}{t^2}\)

But \(s=4.0 \ m\) and \(t=4.0 \ s\) (given)

\(a=\frac{2(4.0\ m)}{(4.0\ s)^2}=\frac{1}{2}\ ms^{-2}\)

Substituting this value of \(a\) in eqn. \((i)\), we get

\(\frac{1}{2}\ ms^{-2} =10\ ms^{-2}\left( \frac{1}{2}-\mu_k \frac{\sqrt3}{2} \right)\)

\(\frac{1}{10}=1-\sqrt3 \mu_k\) or \(\sqrt3 \mu_k=1-\frac{1}{10}=\frac{9}{10}=0.9\)

\(\mu_k =\frac{0.9}{\sqrt3}=0.5\)