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Sagot :
The sled accelerates with magnitude [tex]a[/tex] such that
[tex]\left(2.3\dfrac{\rm m}{\rm s}\right)^2 = 2a(9.2\,\mathrm m) \implies a = 0.2875\dfrac{\rm m}{\mathrm s^2}[/tex]
By Newton's second law, the net force in the plane of motion (parallel to the ground) is
[tex]31\,\mathrm N - F_{\rm friction} = (17 \,\mathrm{kg}) \left(0.2875 \dfrac{\rm m}{\mathrm s^2}\right)[/tex]
so that the force of friction exerts a magnitude of
[tex]F_{\rm friction} = 26.1125 \,\mathrm N[/tex]
Perpendicular to the ground, the sled is in equilibrium, so Newton's second law says
[tex]F_{\rm normal} - (17\,\mathrm{kg})g = 0 \implies F_{\rm normal} = 166.6 \,\mathrm N[/tex]
The magnitude of friction is proportional to the magnitude of the normal force by a factor of [tex]\mu_k[/tex], the coefficient of kinetic friction. It follows that
[tex]F_{\rm friction} = \mu_k F_{\rm normal} \implies \mu_k = \dfrac{26.1125\,\rm N}{166.6\,\rm N} \approx \boxed{0.16}[/tex]
(and the coefficient is dimensionless).
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