Posted by **Anonymous** on Monday, February 21, 2011 at 11:13am.

A snowboarder of mass 74.4 kg (including gear and clothing), starting with a speed of 5.0 m/s, slides down a slope at an angle è = 37.1° with the horizontal. The coefficient of kinetic friction is 0.116. What is the net work done on the snowboarder in the first 6.02 s of descent

- physics -
**Henry**, Tuesday, February 22, 2011 at 3:35pm
Fs = 74.2kg * 9.8 = 727.2N @ 37.1deg = Force of snowboarder.

Fp = 727.2sin37.1 = 438.6N = Force parallel to plane.

Fv = 727.2cos37.1 = 580N = Force perpendicular to plane.

Ff = u*Fv = 0.116 * 580 = 67.3N = Force of friction.

Fn = Fp - Ff = 438.6 - 67.3 = 371.3N = Net force acting on snowboarder.

a = Fn/m = 371.7 / 74.2 = 5m/s^2.

d = Vo*t +0.5at^2,

d = 5*6.02 + 0.5 * 5 * (6.02)^2,

d = 30.1 + 90.6 = 120.7m.

W = Fn * d = 371.3 * 120.7 = 44816 Joules.

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