Question

1. 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? For full credit you must use the work – energy theorem or conservation of energy. (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?

Answer #1

mass,m=85.0 kg

Initaial speed,v_{i} =29.0 m/s

Final speed,v_{f}= 1.95 m/s

work done,W= 3995.0 J

Vertical height we need to find out first. Using work energy theorm we get,

(1/2)*mv_{i}^{2} + mgyi =
(1/2)*m*v_{f}^{2} + mgh + 3995.0 J,

where the initial y value is zero, the initial speed is 29.0 m/s and the final speed 1.95 m/s.

So by substituting we get, h = 37.9 m.

(b)Work done is, Wg = -mgh = = -31600 J. This work is negative since the the gravitational force acts downward and the vertical component of displacement points up in the opposite direction.

(c) ZERO since the normal force and motion are perpendicular

(d) L moved along the incline is given as,L = h/sin 30 = 75.8 (m)

(e)force of friction between the skis and snow is constant with magnitude f.= -3995.0 J = -f*L, where L = 78.5 (m); thus f = 52.7(N).

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