The figure below shows the free body diagram of the block.
The friction force acting on the block is given as,
f=µk(mgcosθ) The force equation for the block along the plane of inclination,
mgsinθ−f=ma mgsinθ−µk(mgcosθ)=ma a=10×0.6−×10×0.8 a=5m∕s2 The distance
s is given from
∆EBC EC=s===m The velocity at final position is give as,
v2=2as Substituting the values in above expression, we get
v2=2×5m∕s2×m v=m∕s The component of velocity at point
C is given as,
Using second equation of motion, we get
h=ut+gt2 10=utsinθ+gt2 5t2+5t−10=0 t2+t−2=0 On solving the time comes out to be
t=1 s The distance DF is,
DF=vcosθ=×0.8=m