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?

1 Answer
Apr 15, 2018

d=52m

Explanation:

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

Ff=Fnμ

For that we need the force normal which is equivalent to the Force of weight in this case:
Fn=mg
Ff=mgμ
Ff=(12.4kg)(9.81ms2)(0.35)
Ff=42.575N

So then the Net force is:
Fnet=74N42.575N

Fnet=31.425N

Using the net force I can determine the net acceleration:
Fnetm=anet=31.425N12.4kg=2.534ms2

Using that:
d=vit+12at2
Since vi=0ms
d=12at2
d=12(2.534ms2)(6.4s)252m