Question

please help!! (include steps)

At the 1984 Winter Olympics, William Johnson of the United States reached a speed of 104.5 km/h in the downhill skiing competition. Suppose Johnson left the slope at that speed and then slid freely along a horizontal surface. If the coefficient of kinetic friction between the skis and the snow was 0.120 and his final speed was half of his initial speed, find the distance William traveled.

Answer 1

Hi there!

Begin by converting 104.5 km/h to m/s.

$\frac{104.5km}{hr} * \frac{hr}{3600 s}* \frac{1000m}{ km} = 29.028 m/s$

Recall the definition of work:

$\large\boxed{W = \Delta KE = Fdcos\theta}}$

AND:

$\large\boxed{W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The work done in this situation is due to the friction force:

$F = \mu mg\\\\W = \mu mgd(cos180) \\W = -\mu mgd$

Now, using the change in kinetic energy, we can solve:

$-\mu mgd = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Cancel out the mass.

$-\mu gd = \frac{1}{2}v_f^2 - \frac{1}{2}v_i^2$

Rearrange for a working equation:

$d = \frac{\frac{1}{2}v_f^2 - \frac{1}{2}v_i^2}{-\mu g}$

Plug in the given values:

$d = \frac{\frac{1}{2}(14.51)^2 - \frac{1}{2}(29.028)^2}{-(.120)(9.8)}$

Solve:

$d = \frac{\frac{1}{2}(14.51)^2 - \frac{1}{2}(29.028)^2}{-(.120)(9.8)} = \boxed{268.74 m}$