If a #1/2 kg# object moving at #7/4 m/s# slows to a halt after moving #3/8 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Jan 21, 2017

Answer:

The answer is #mu_k=1.25#

Explanation:

We solve in the horizontal direction #rarr^+#

#u=7/4ms^-2#

#v=0#

#s=3/8m#

#a=# acceleration

We use the equation

#v^2=u^2+2as#

#0=49/16+2*3/8*a#

#a=-49/16*4/3=-49/12ms^-2#

By Newton`s second Law

#F_r=ma=-49/4*1/2=-49/8N#

The reaction is #N=1/2g=1/2*9.8=4.9N#

The coefficient of kinetic friction is #mu_k=F_r/N=49/8*1/4.9=1.25#