If an object is moving at #7# #ms^-1# over a surface with a kinetic friction coefficient of #u_k=14 /g#, how far will the object continue to move?

1 Answer
Feb 17, 2016

Answer:

Finding the frictional force and the acceleration due to it allows us to find the distance traveled, #1.75# #m#, before the object stops.

Explanation:

The force accelerating (decelerating) the mass will be the frictional force, which is the frictional coefficient times the normal force:

#F_"fric"=muF_N#

In this case the normal force is the weight force of the object:

#F_N=mg#

So

#F_"fric"=mumg=14/g*m*g#

#g# cancels out to leave a frictional force of #14m# where #m# is the mass. We don't know it, but be patient. ;-)

The acceleration (deceleration) of the object with this force acting on it is given by Newton's Second Law :

#a=F/m=(14m)/m=14# #ms^-2# (#m# cancels)

This should be given a minus sign, because it is an acceleration in the opposite direction to the object's velocity - a deceleration.

#a=-14# #ms^-2#

We know the initial velocity #u=7# #ms^-1# and the final velocity #v=0# #ms^-1# and the acceleration, and we are asked for the distance the object moves before stopping.

#v^2=u^2+2ad#

Rearranging:

#d=(v^2-u^2)/(2a)=(0^2-7^2)/(2*-14)=49/28 = 1.75 m#