91Ó°ÊÓ

New ski is 200 m slope in the Alps, a skier reduced his top-to-bottom time from 61 s with standard skis to 42 s .

Frictional force is given by the product of coefficient of friction and the normal reaction. Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass. Using these concepts, the problem can be solved.

Formula:

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{mg}\mathbf{}\mathbf{²õ¾\pm ²Ôθ}\mathbf{-}{\mathbf{f}}_{\mathbf{k}}$

(a)

If the distance is L during time t with initial speed zero and a constant acceleration a, then

$L=a{t}^2/2,$

Which gives the acceleration for the first (old) pair of skis:

$$\begin{gathered} a_1=\frac{2L}{t_1^2} \\ =\frac{2\left(200m\right)}{{\left(61s\right)}^2} \\ =0.11m/s^2 \end{gathered}$$

(b)

The acceleration for the second (new) pair is

$$\begin{gathered} a_2=\frac{2L}{t_1^2} \\ =\frac{2\left(200m\right)}{{\left(42s\right)}^2} \\ =0.23m/s^2 \end{gathered}$$

(c)

The net force along the slope acting on the skier of mass m is

$$\begin{gathered} F_{net}=mg\sin \mathrm{θ}-f_k \\ =mg\left(\sin \mathrm{θ}-{\mathrm{μ}}_k\cos \mathrm{θ}\right) \\ =ma \end{gathered}$$

Which we solve for for the first pair of skis:

$$\begin{gathered} {\mathrm{μ}}_{k1}=\tan \mathrm{θ}-\frac{a_1}{g\cos \mathrm{θ}} \\ =\tan \left(3°\right)-\frac{0.11m/s^2}{\left(9.8m/s^2\right)\cos \left(3°\right)} \\ =0.041 \end{gathered}$$

(d)

For second pair, we have

$$\begin{gathered} {\mathrm{μ}}_{k2}=\tan \mathrm{θ}-\frac{a_2}{g\cos \mathrm{θ}} \\ =\tan \left(3°\right)-\frac{0.23m/s^2}{\left(9.8m/s^2\right)\cos \left(3°\right)} \\ =0.029 \end{gathered}$$