If a #13 kg# object moving at #7 m/s# 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
Jan 9, 2016

Answer:

#mu_k=0.25#

Explanation:

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

1. Method 1 - Using Newton's Laws and Equations of Motion :

The acceleration of the object is uniform and in 1 direction and can be found from a relevant equation of motion for constant acceleration in 1 dimension as follows :

#v^2=u^2+2ax#

#therefore a=(v^2-u^2)/(2x)=(0^2-7^2)/(2xx10)=-2.45m//s^2#.

This is acceleration is caused by the frictional force which is the resultant force acting on the object and so by Newton 2 and definition of frictional forces we get :

#sumvecF=mveca#

#therefore -f_k=ma#, where #f_k=mu_kN#.

But #N=mg =>mu_kmg=ma#.

#therefore mu_k=a/g=2.45/9.8=0.25#.

2. Method 2 - Using energy considerations :

The Work-Energy Theorem states that the work done by the resultant force equals the change in kinetic energy brought about.

#therefore W_(f_k)=DeltaE_k#

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

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

#therefore mu_k=(v^2-u^2)/(2gx)=(0^2-7^2)/(2xx9.8xx10)=0.25#.