If a #2# #kg# object moving at #6# #ms^-1# slows down to a halt after moving #10# #m#, what is the friction coefficient of the surface that the object was moving over?

1 Answer
Apr 28, 2016

Answer:

The object's initial kinetic energy will be used as work overcoming the frictional force. The calculation below shows that the coefficient of friction is approximately 5.4.

Explanation:

The object's initial kinetic energy is given by:

#E_k=1/2mv^2=1/2xx2xx6^2=36# #J#

This energy will be dissipated doing work against the frictional force acting on the object.

#W=Fd#

Rearranging:

#F=W/d=(36J)/(10m)=3.6# #N#

This is the magnitude of the frictional force, which is given by:

#F_"frict"=muF_N#, where #F_N# is the normal force, which in turn is the weight force of the object:

#F_N=mg=2*9.8=19.6# #N#

To find the frictional force, then:

#mu=(F_N)/(F_"frict")=19.6/3.6~~5.4#

(coefficients of friction do not have units)

(this is an improbable value for a coefficient of friction - they are typically less than 1, but occasionally as high as 2. Teachers and textbooks do sometimes write questions with these improbable values, though)