Question

A force of 100.0N accelerates a 5.0 kg box at 15.0 m/s^2 along a level surface. Find the coefficient of sliding friction for the box. After that, how far will the box have gone after 5.0 seconds of pulling it?

Answer 1

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 :

`d = ut + (at^2)/(2)\\\\d = 0 + (15* 5^2)/(2)\\\\d = 187.5\ m`

Hence, this is the required solution.