Here, The coefficient of friction is given as, µ=Cs2 The figure below shows the free body diagram of the block on an inclined plane.
Balancing forces on the block we get, Mg‌sin‌θ−f=Ma Mg‌sin‌θ−µMg‌cos‌θ=Ma g‌sin‌θ−µg‌cos‌θ=a g(sin‌θ−Cs2‌cos‌θ)=a And, the acceleration is given as, a=
dv
dt
=
dv
ds
.
ds
dt
ads = vdv Substituting the value of a in above we get, g(sin‌θ−Cs2‌cos‌θ)ds=vdv Integrate both sides. (g‌sin‌θ)s−g
Cs3
3
‌cos‌θ=
v2
2
+K For θ=45° (
g
√2
)s−(
Cs3
3
)g√2=
v2
2
+K Initially, t=0,s=0,u=0 The constant K comes out to be 0. So, (