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 angle 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?

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wagonbelleville wagonbelleville · Answer 1 · 2020-02-11T09:52:13+01:00

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

[tex]m g sin\theta - F_{rope} = ma[/tex]

velocity is constant, a = 0

[tex]m g sin\theta - F_{rope} =0[/tex]

[tex]F_{rope} = m g sin\theta[/tex]

[tex]F_{rope} = 70.1\times 9.8\times sin 8.6^0[/tex]

[tex]F_{rope}= 102.73\ N[/tex]

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

[tex]F_{rope}-m g sin\theta = ma[/tex]

velocity is constant, a = 0.135 m/s₂

[tex]F_{rope} = m g sin\theta+ma[/tex]

[tex]F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135[/tex]

[tex]F_{rope}= 112.19\ N[/tex]