Question

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?

Answer 1

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* 9.8* 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* 9.8* sin 8.6^0+70.1* 0.135`

`F_(rope)= 112.19\ N`