A 85.0-kg speed skier has finished a long down hill race and reaches a final slo

Question

A 85.0-kg speed skier has finished a long down hill race and reaches a final slope (fig. 1 below) designed to slow her down. At the bottom of this slope her speed is 29.0 m/s. She slides up the inclined plane of snow on her skis and at a certain vertical height h has speed 1.95 m/s. The force of friction between her skis and the snow does work of magnitude 3995.0 J . (Ignore air friction.) (a) What is the vertical height h? (b) What is the work Wg done by gravity during her trip up the inclined plane? Is this work positive or negative ? Give a brief explanation of the correct sign you choose (positive or negative.) (c) What is the work done by the normal force ? Explain your answer. (d)Use basic trigonometry to find the distance L moved along the incline. See fig. 2 for reference. (e) Assume the force of friction between the skis and snow is constant with magnitude f. What is f?

Explanation / Answer

a ) Net work done = Change in KE Wg + Wf = 0.5 * m * (v2^2 - v1^2) Since the displacement in up the incline...and both the friction and gravity opposes motion, hence both these work will be negative -85 * 9.81 * h -3995 = 0.5 * 85 * (1.95 ^2 - 29 ^2 ) so...h = 37.88 m b) Work done by gravitational force = -m*g*h = -31585.89 J This work is negative as the force acts opposite to the direction of motion c) Work done by normal force will be zero reason : It acts perpendicular to the incline...and all the motion is parallel to the incline...so normal force does not cause any work done. d) sin {30} = h / L so, L = h / sin 30 = 75.76 m e ) Force of friction = Work done / distance = 3995 / 75.76 = 52.732 N

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