7. The free-body diagram for the puck is shown below.
Nis the normal force of the ice on the puck,
fis
the force of friction (in the −xdirection), and mgis the force of gravity.
(a) The horizontal component of Newton’s second law gives −f=ma, and constant acceleration
kinematics (Table 2-1) can be used to find the acceleration.
Since the final velocity is zero, v2=
v2
0+2ax leads to a=−v2
0/2x.Thisis
substituted into the Newton’s law equa-
tion to obtain
f=mv2
0
2x
=(0.110 kg)(6.0m/s)2
2(15 m)
=0.13 N .
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N
mg
f
(b) The vertical component of Newton’s second law gives N−mg =0,soN=mg which implies (using
Eq. 6-2) f=µkmg. We solve for the coefficient:
µk=f
mg =0.13 N
(0.110 kg) 9.8m/s2=0.12 .