If an object is moving at #7# #ms^-1# over a surface with a kinetic friction coefficient of #u_k=2/g#, how far will the object continue to move?

Question

Layla Ahmed · Accepted Answer

The distance traveled is #24.5# #m#.

The mass cancels out and so does the value of 'g', and we can use the calculated acceleration and the fact that the final velocity is 0 to find the distance traveled.

The force causing the deceleration will be the frictional force, given by #F_"frict"=muF_"norm"#, where the normal force #F_"norm"=mg#, so #F_"frict"=mumg#. The acceleration of the object will be given by #a=F_"frict"/m=(mumg)/m# The mass cancels, which is convenient since we have not been told its value: #a=mug=2/g xx g# The g also cancels, so that the acceleration is just equal to #2# #ms^-2#. Since it is in the opposite direction to the initial velocity, we can write this as #-2# #ms^-2#. Now we can use #v^2=u^2+2as#. The object comes to rest, so the final velocity, #v=0# #ms^-1#. #0^2 = 7^2 +(-2)s# Organizing: #s=49/2=24.5# #m#

Sebastian Acosta · Answer

undefined To find the distance (d) the object will continue to move, you can use the formula:

$$ d = \frac{v^2}{2 \cdot u_k \cdot g} $$

where:
$ v $ = initial velocity (7 m/s)
$ u_k $ = kinetic friction coefficient ($ \frac{2}{g} $)
$ g $ = acceleration due to gravity (approximately 9.8 m/s²)