If a #3/2 kg# object moving at #3/4# #ms^-1# slows to a halt after moving #5/8 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Aug 1, 2016

Answer:

First find the acceleration, then use it to find the magnitude of the frictional force, and use that to find the coefficient of friction.

Explanation:

First find the acceleration:

#v^2 = u^2 + 2as#

Rearranging to make #a# the subject:

#a = (v^2-u^2)/(2s)#

#a = (0^2-(3/4)^2)/(2xx(5/8)) = -0.45# #ms^-2#

The negative sign just shows that the object is decelerating - its acceleration is in the opposite direction to the initial velocity.

Now to find the frictional force:

#F = ma = (3/2)xx(0.45) = 0.675# #N#

The frictional force is given by #F_"frict" = muF_"norm"#, where the normal force is #F_"norm"=mg#.

Rearranging to make the frictional coefficient the subject:

#mu=(F_"frict")/(F_"norm") = 0.675/(3/2xx9.8) = 0.022#

(frictional coefficients have no units, since they are ratios of forces)

(note: technically, a coefficient of friction is between two bodies, not a property of the surface alone)