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arsen [322]
3 years ago
10

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

e of 8.6° with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 1.78 m/s and (b) v = 1.78 m/s as v increases at a rate of 0.135 m/s2?
Physics
1 answer:
lara [203]3 years ago
7 0

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

   m g sin\theta - F_{rope} = ma

 velocity is constant, a = 0

   m g sin\theta - F_{rope} =0

   F_{rope} = m g sin\theta

   F_{rope} = 70.1\times 9.8\times sin 8.6^0

  F_{rope}= 102.73\ N

b) now, when acceleration, a = 0.135 m/s²

   F_{rope}-m g sin\theta = ma

 velocity is constant, a = 0.135 m/s₂

   F_{rope} = m g sin\theta+ma

   F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135

  F_{rope}= 112.19\ N

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