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
Hriman
Apr 15, 2018

#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#

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