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=(FpFk)/m=(510.172.55)/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 = (510.172.55) * 210.4 =
92,061 Joules.
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