Answer:
a). Acceleration with old model [tex]a_{o}=8.343\frac{m}{s^{2}}[/tex]
b). Acceleration with new model [tex]a_{n}=11.785\frac{m}{s^{2}}[/tex]
c). Coefficient of kinetic friction old model [tex]u_{k}=0.0515[/tex]
d). Coefficient of kinetic friction old model [tex]u_{k}=0.0464[/tex]
Explanation:
Using the Newton laws so the initial speed is zero
so:
Δ[tex]x=v_{o}*t+\frac{1}{2}*a*t^{2}[/tex]
[tex]v_{o}[/tex] because the ski began from the rest initial velocity is zero
Δ[tex]x=\frac{1}{2}*a*t^{2}[/tex]
a).
[tex]a=\frac{2*x}{t^{2}}[/tex]
[tex]a_{o}=\frac{2*200m}{(47.94s)^{2}}[/tex]
[tex]a_{o}=8.343\frac{m}{s^{2}}[/tex]
b).
[tex]a=\frac{2*x}{t^{2}}[/tex]
[tex]a_{n}=\frac{2*200m}{(33.94s)^{2}}[/tex]
[tex]a_{n}=11.785\frac{m}{s^{2}}[/tex]
In the horizontal direction the friction force acts on the ski is:
[tex]f_{k}=u_{k}*m*g*cos(\alpha)[/tex]
[tex]u_{k}=\frac{m*g*sin(\alpha)-m*a}{m*g*cos(\alpha)}[/tex]
c).
[tex]u_{k}=\frac{g*sin(\alpha)-a}{g*cos(\alpha)}[/tex]
[tex]u_{k}=\frac{9.8*sin(4)-0.108}{9.8*cos(4)}[/tex]
[tex]u_{k}=0.0515[/tex]
d).
[tex]u_{k}=\frac{g*sin(\alpha)-a}{g*cos(\alpha)}[/tex]
[tex]u_{k}=\frac{9.8*sin(4)-0.23}{9.8*cos(4)}[/tex]
[tex]u_{k}=0.0464[/tex]
