If a #3 kg# object moving at #5 m/s# slows to a halt after moving #10 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Dec 15, 2015

Answer:

#mu_k=0,12755#

Explanation:

I will show you 2 different methods to do this question :

Method 1 - Using Newton's Laws and equations of motion

The acceleration of the object may be found from the equations of motion for constant linear acceleration as follows :
#v^2=u^2+2ax#
#thereforea=(v^2-u^2)/(2x)=(0^2-5^2)/(2xx10)=-1,25m//s^2#

This acceleration is caused by the resultant force of friction acting on the object and by Newton 2 given as :
#sumF=ma#
#therefore -f_k=ma#
#therefore -mu_kN=ma#
#therefore -mu_kmg=ma#

#therefore mu_k=(ma)/(-mg)=(-1,25)/(-9,8)=0,12755#.

Method 2 - Using energy considerations

From conservation of energy, work done by friction equals change in kinetic energy brought about.

#therefore W_ ( fk ) = Delta E_K#

#therefore f_k*x=1/2m(v^2-u^2)#

#therefore mu_kmg*x=1/2mv^2#

#therefore mu_k=(1/2mv^2)/(mgx)=(1/2xx5^2)/(9,8xx10)=0,12755#.