A 51.7 kg skier is moving at 12.5 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 1.7 m long. The coefficient of kinetic friction between this patch and her skis is 0.29. After crossing the rough patch, what is the speed of the skier in [tex]\text{m/s}[/tex]? Assume [tex]g = 9.8 \, \text{m/s}^2[/tex].

The final speed of the skier, after crossing the 1.7 m rough patch with a kinetic friction coefficient of 0.29, will be 3.11 m/s. This is due to the energy lost to friction, slowing her down.

Explanation:

This is a physics problem that incorporates principles of kinetics and friction. The 51.7 kg skier is initially moving at 12.5 m/s across a frictionless surface. Once she hits the patch of rough, frictional snow that is 1.7 m long, some of her kinetic energy will be converted into heat due to friction, slowing her down. The force of friction (F) is calculated as mass (m) x gravity (g) x coefficient of kinetic friction (µk), so F = 51.7 kg x 9.8 m/s² x 0.29 = 147.1 N. This force acts over the rough patch (dF), doing work on the skier and slowing her down. Work done (W) is Force (F) x the distance over which it's applied (d), so W = 147.1 N x 1.7 m = 250.07 J.

Knowing that work done by friction equals the change in kinetic energy, we can set up the equation 1/2 x mass x velocity² = Work done, or 1/2 x 51.7 kg x (v_final)² = 250.07 J. Solving for final velocity (v_final), we end up with v_final = √((2 x 250.07 J) / 51.7 kg) = √9.67 = 3.11 m/s. So the skier will be moving at 3.11 m/s after crossing the rough patch.

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