Question

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?

Answer 1

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`