By balancing the force, we get :
100 - Frictional Force = ma
100 - μmg = ma
100 - μ(5)(10) = (5)(15)
50μ = 25
μ = 0.5
Distance travelled by box in 5 seconds is :
[tex]d = ut + \dfrac{at^2}{2}\\\\d = 0 + \dfrac{15\times 5^2}{2}\\\\d = 187.5\ m[/tex]
Hence, this is the required solution.
