If an object with a mass of #10 kg # is moving on a surface at #15 ms^-1# and slows to a halt after # 4 s#, what is the friction coefficient of the surface?

1 Answer
Jan 20, 2016

Answer:

The acceleration of the object is #3.75 ms^-2#, which means the frictional force acting is #37.5 N# and the frictional coefficient is #mu = 0.38#.

Explanation:

First we calculate the acceleration of the object:

#v=u+at#

where #v# is final velocity #(ms^-1)#, #u# is initial velocity #9ms^-1)#, #a# is the acceleration #(ms^-2)# and #t# is the time #(s)#.

Rearranging:

#a=(v-u)/t=(0-15)/4 = -3.75 ms^-2#

The negative sign just shows that it is a deceleration, and we can ignore it for the rest of this calculation.

Now we find the force acting to slow the object, using Newton's Second Law :

#F=ma=10*3.75=37.5 N#

This is simply the frictional force acting. Frictional force is related to the normal force acting on the object, which in this case is the weight of the object, #F_N=mg=10*9.8 = 98 N#:

#F_f = muF_N#

Rearranging and substituting in the force:

#mu = F_f/F_N = 37.5/98 = 0.38#

Note that #mu# is a dimensionless number, that is, it doesn't have any units.