Physics, 26.08.2019 04:50 sadielebaron

A60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees. find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (hint: find the distance traveled up the incline assuming a straight-line path as shown in the figure.)

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Answer from: GracieMaeB

Assuming a straight-line path as shown in the figure, the skier's final speed at the top is 9.5 m/s.

Given the following data:

Mass of skier = 60 kgInitial speed = 12.0 m/sHeight = 2.50Coefficient of friction = 0.08

To find the skier's final speed at the top:

First of all, we would determine the distance traveled up the incline (assuming a straight-line path) by using trigonometry:

From Sine law:

$Distance = \frac{h}{sin \theta} \\\\Distance = \frac{2.5}{sin (35)} \\\\Distance = \frac{2.5}{0.5736}$

Distance, d = 4.36 meters.

The initial kinetic energy possessed by the skier would generate a final kinetic energy, potential energy and heat due to friction as work is done.

Mathematically, this given by the formula:

K.E_i = K.E_f + P.E + Work

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_f^2 + mgh + Fd$ .....equation 1.

Since we're solving for her final speed, we would rearrange the formula:

$\frac{1}{2} mv_f^2 = \frac{1}{2} mv_i^2 - mgh - Fd$ .....equation 2.

We would apply Newton's Second Law of Motion to determine the force acting on the skier as she ascend straight-line path:

$F = umgcos\theta\\\\F = 0.08(60)(9.8)cos(35)\\\\F = 47.04(0.8192)$

Force, F = 38.54 Newton.

Substituting the values into eqn 2, we have:

$\frac{1}{2} (60)v_f^2 = \frac{1}{2} (60)(12)^2 - 60(9.8)(2.50) - 38.54(4.36)\\\\30v_f^2 = 4320 - 1470 - 168.03\\\\30v_f^2 = 4320 - 1638.03\\\\30v_f^2 = 2681.97\\\\v_f^2 = \frac{2681.97}{30}\\\\v_f^2 = 89.4\\\\v_f = \sqrt{89.4}$

Final speed, v_f = 9.5 m/s

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A60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees...

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