Posted by **JAY Z** on Monday, December 2, 2013 at 2:58pm.

A snowboarder of mass 87.1 kg (including all of the gear and clothing), starting with a downhill speed of 6.1 m/s, slides down a slope at an angle $\theta$ of 36.7° with the horizontal. The coefficient of kinetic friction is 0.106. What is the net work done on the snowboarder in the first 8.02 s of descent?

- Physics -
**Henry**, Tuesday, December 3, 2013 at 10:18pm
Fs = m*g = 87.1kg * 9.8N/kg = 853.6 N. =

Force of snowboarder.

Fp = 853.6*sin36.7 = 510.1 N. = Force parallel to hill.

Fn = 853.6*cos36.7 = 684.4 N = Normal =

Force perpendicular to hill.

Fk = u*Fn = 0.106 * 684.4 = 72.55 N. =

Force of kinetic friction.

a=(Fp-Fk)/m=(510.1-72.55)/87.1=5.02 m/s^2.

d = Vo*t + 0.5a*t^2

d = 6.1*8.02 + 2.51*8.02^2 = 210.4 m.

Down the slope.

Work = F*d = (510.1-72.55) * 210.4 =

92,061 Joules.

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