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Question #86690
A dog (mass 11.7 kg) sits on a sled (mass 5.1 kg) that is being pushed with a force of 35 N on a flat snowy surface. If the acceleration of both is 1.7 m/s2, what is the coefficient of kinetic friction for the sled sliding on the snow?
Expert's answer
The Newton's second law states
m
a
=
F
−
F
f
r
i
c
ma=F-F_{\rm{fric}}
ma
=
F
−
F
fric
(
11.7
+
5.1
)
1.7
=
35
−
F
f
r
i
c
(11.7+5.1)1.7=35-F_{\rm{fric}}
(
11.7
+
5.1
)
1.7
=
35
−
F
fric
So, the friction force
F
f
r
i
c
=
6.44
N
F_{\rm{fric}}=6.44\:\rm{N}
F
fric
=
6.44
N
The coefficient of kinetic friction
μ
k
=
F
f
r
i
c
m
g
=
6.44
(
11.7
+
5.1
)
9.81
=
0.04
\mu_k=\frac{F_{\rm{fric}}}{mg}=\frac{6.44}{(11.7+5.1)9.81}=0.04
μ
k
=
m
g
F
fric
=
(
11.7
+
5.1
)
9.81
6.44
=
0.04
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