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August 31, 2015

# Homework Help: physics

Posted by JAYZ on Friday, November 22, 2013 at 1:59pm.

A skier of mass 77.7 kg, starting from rest, slides down a slope at an angle $\theta$ of 35.7° with the horizontal. The coefficient of kinetic friction is 0.08. What is the net work, i.e. net gain in kinetic energy, (in J) done on the skier in the first 10.9 s of descent?
• physics - Henry , Friday, November 22, 2013 at 8:27pm

Ws = m*g = 77.7kg * 9.8N/kg = 761.5 N = Wt. of skier.

Fp = 761.5*sin35.7 = 444.3 N. = Force parallel to slope.

Fn = 761.5*cos35.7 = 618.4 = Normal = Force perpendicular to slope.

Fk = u*Fn = 0.08 * 618.4 = 49.47 N. =
Force of kinetic friction.

Fp-Fk = m*a
a=(Fp-Fk)/m=(444.3-49.47)/77.7=5.08m/s^2
V = Vo + a*t
V = 0 + 5.08*10.9 = 55.39 m/s.

Work=m*V^2/2 = 77.7*55.39^2/2=119,194 J.

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