A skier is being pulled up a smooth 25° dry ski slope by a rope which makes an angle of 35° with the horizontal. The mass of the skier is 75 kg and the tension in the rope is 350 N. Initially the skier is at rest at the bottom of the slope. The slope is smooth. Find the skier’s speed after 5 s and

Question

A skier is being pulled up a smooth 25° dry ski slope by a rope which makes an angle of 35° with the horizontal. The mass of the skier is 75 kg and the tension in the rope is 350 N. Initially the skier is at rest at the bottom of the slope. The slope is smooth. Find the skier’s speed after 5 s and find the distance he has travelled in that time.

Accepted Answer

In the diagram the skier is modelled as a particle. Since the skier moves parallel to the slope consider motion in that direction.

Resultant force = mass × acceleration

350 cos 10° - 75g sin 25° = 75 × a

[latex]\begin{matrix}\boxed{ ext{Taking g as 9.8 }} & \longrightarrow a = \frac{34.06}{75} = 0.454 ( ext{to 3 d.p.}).\end{matrix}[/latex]

This is a constant acceleration so use the constant acceleration formulae.

v = u + at

[latex]\begin{matrix} v = 0 + 0.454 imes 5 & \longleftarrow \boxed{u = 0, a = 0.454, t = 5} \end{matrix} [/latex]

Speed = 2.27 [latex]ms^{-1}[/latex] ( to 2 d.p.).

[latex]s = ut + \frac{1}{2}at^{2}[/latex]

[latex]s = 0 + \frac{1}{2} × 0.454 × 25[/latex]

Distance travelled = 5.68 m (to 2 d.p.).

Question 5.5: A skier is being pulled up a smooth 25° dry ski slope by a r...

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