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

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Mar 13, 2018

Answer:

#mu_k = 0.002548#

Explanation:

Well, the force of friction #F_f# is defined as the following...

#F_f = mu_k N#

And the normal is given by...

#N = mg#

#N = (2)(9.81)#

#N = 19.62N#

So now, we need to calculate an acceleration to get the force of friction. We are assuming in this question that #F_(drive) = F_f# because all the energy that the mass possess is lost to friction.

#v^2=u^2-2as#

#0^2 = 0.5^2 -2(5)a#

#0 = 0.25-10a#

#a = (-0.25)/(10)#

#a= -0.025 ms^-2#

#F_f=ma#

#F_f=(1)abs(-0.025)#

#F_f=0.025N#

#mu_k = F_f/N#

#mu_k = 0.025/9.81#

#mu_k = 0.002548#

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