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A 55-kg person on skis starts from rest down a hill sloped at 35Ā° from the horizontal. The coefficient of friction between the skis and the snow is 0.12. After the skier had been moving for 5.0 s, the friction of the snow suddenly increased and made the net force on the skier zero. What is the new coefficient of friction?

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Answer:

k = 0.7

Explanation:

mass (m) = 55 kg

slope of hill (Īø) = 35 degrees

coefficient of friction (k) = 0.12

time (t) = 5 s

acceleration due to gravity (g) = 9.8 m/s^{2}

What is the new coefficient of friction if the friction of the snow suddenly increased and made the net force on the skier zero?

we can find the coefficient of friction by applying newtons second law of motion, force = mass x acceleration

where

  • force = force of the skier - friction force
  • force of the skier = mg.SinĪø
  • friction force = k.mg.CosĪø
  • therefore force = mg.SinĪø - k.mg.CosĪø

therefore force = mass x acceleration now becomes

mg.SinĪø - k.mg.CosĪø = mass x acceleration

mg.SinĪø - k.mg.CosĪø = ma

mg (SinĪø - k.CosĪø) = ma

g(SinĪø - k.CosĪø) = a

for the net force on the skier to be zero, the acceleration of the skier has to be zero.

therefore g(SinĪø - k.CosĪø) = 0

now substituting the required values into the equation above we have

9.8 ( sin 35 - k.cos 35 ) = 0

9.8 (0.5736 - k.0.8192) = 0

5.6213 - 8.028k = 0

5.6213 = 8.028k

k = 0.7

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