Question

A 12.4kg mass is being pushed, from rest, along a level surface. The coefficient of kinetic friction between the mass and the surface is 0.35. The mass is being pushed with a force of 74N [North]. The force is being exerted for 6.4 s. ?

What is the total distance that the mass will travel, before it comes to a rest?

Answer 1

`d= 52 m`

Explanation:

First thing I need is to figure out the Force of Friction:

`F_f= F_n*mu`

For that we need the force normal which is equivalent to the Force of weight in this case:
`F_n= mg`
`Ff= mgmu`
`F_f= (12.4 kg)(-9.81 m/s^2)(0.35)`
`F_f=-42.575 N`

So then the Net force is:
`F_("net")= 74 N-42.575 N`

`F_("net")= 31.425 N`

Using the net force I can determine the net acceleration:
`F_("net")/m=a_("net")= (31.425 N)/(12.4 kg)= 2.534 m/s^2`

Using that:
`d= v_it+1/2at^2`
Since `v_i= 0m/s`
`d= 1/2at^2`
`d= 1/2(2.534 m/s^2)(6.4s)^2 approx 52 m`