Given,
The mass of the skater, m=63 kg
The coefficient of static friction, μs=0.4
The coefficient of the kinetic friction, μk=0.02
F₁=242 N
F₂=162 N
(c) The static friction is given by,
[tex]f_s=N\mu_s[/tex]Where N is the normal force.
The normal force acting on the skater is
[tex]N=mg[/tex]Where g is the acceleration due to gravity.
Therefore the static friction is given by,
[tex]f_s=mg\mu_s[/tex]On substituting the known values,
[tex]\begin{gathered} f_s=63\times9.8\times0.4 \\ =246.96\text{ N} \end{gathered}[/tex]Therefore the static friction on the skater is 246.96 N
d)The net force acting on the skater is
[tex]\begin{gathered} F_{\text{net}}=ma_{} \\ =F_{\text{tot}}-f \\ =F_{\text{tot}}-N\mu_k \\ =F_{\text{tot}}-mg\mu_k \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} 63a=291.2-63\times9.8\times0.02 \\ a=\frac{278.85}{63} \\ =4.43m/s^2 \end{gathered}[/tex]Thus the acceleration of the skater is 4.43 m/s²