Unformatted text preview: 38. This problem involves Newton’s second law for motion along the slope. (a) The force along the slope is given by Fg = mg sin θ − μ FN = mg sin θ − μ mg cos θ = mg (sin θ − μ cos θ ) = (85.0 kg)(9.80 m/s 2 ) [sin 40.0° − (0.04000) cos 40.0°] = 510 N. Thus, the terminal speed of the skier is vt = 2 Fg Cρ A = 2(510 N) = 66.0 m/s. (0.150)(1.20 kg/m3 )(1.30 m 2 ) (b) Differentiating vt with respect to C, we obtain dvt = − 1 2 Fg −3 / 2 1 2(510 N) (0.150) −3/ 2 dC C dC = − 3 2 2 ρA 2 (1.20 kg/m )(1.30 m ) = −(2.20 ×102 m/s)dC. ...
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Force

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