Question

1. a 49 kg snowboarder descends a 0.45 km ski slope from the top of a mountain to the base as shown in the system diagram. if the friction on the slope is negligible and they started at a speed of 2.1 m/s,

a) calculate the speed of the snowboarder when they reach the base of the mountain using energy.

b) if there is a coefficient of friction between the snowboarder and the surface of 0.44 while traveling along the horizontal at the base of the mountain, using energy determine the distance the snowboarder will travel before stopping using energy.

Explanation:

Step1: Find the height of the slope

The length of the slope \(L = 0.45\space km=450\space m\) and the angle \(\theta = 14^{\circ}\). Using the trigonometric relation \(h = L\sin\theta\), we have \(h = 450\times\sin(14^{\circ})\).

$$h = 450\times0.2419\approx108.86\space m$$

Step2: Apply the conservation of mechanical energy (for part a)

The initial mechanical energy \(E_{i}=K_{i}+U_{i}\), where \(K_{i}=\frac{1}{2}mv_{i}^{2}\) (\(v_{i} = 2.1\space m/s\), \(m = 49\space kg\)) and \(U_{i}=mgh\). The final mechanical energy \(E_{f}=K_{f}=\frac{1}{2}mv_{f}^{2}\) (since \(U_{f} = 0\) at the base).
By conservation of energy \(E_{i}=E_{f}\), so \(\frac{1}{2}mv_{i}^{2}+mgh=\frac{1}{2}mv_{f}^{2}\).
Divide through by \(m\): \(\frac{1}{2}v_{i}^{2}+gh=\frac{1}{2}v_{f}^{2}\).
Substitute \(v_{i} = 2.1\space m/s\), \(g = 9.8\space m/s^{2}\), and \(h\approx108.86\space m\)

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Step3: Analyze the energy for part b

The initial kinetic energy at the base (from part a, \(K_{i}=\frac{1}{2}mv_{f}^{2}\), where \(v_{f}\approx46.24\space m/s\)) is dissipated by the work - done against friction. The frictional force \(F_{f}=\mu_{k}N\), and on a horizontal surface \(N = mg\), so \(F_{f}=\mu_{k}mg\). The work - done by friction \(W = F_{f}d=\mu_{k}mgd\), and by the work - energy theorem \(K_{i}=W\) (since \(K_{f} = 0\)).
\(\frac{1}{2}mv_{f}^{2}=\mu_{k}mgd\).
Cancel out \(m\): \(\frac{1}{2}v_{f}^{2}=\mu_{k}gd\).
Substitute \(v_{f}\approx46.24\space m/s\), \(\mu_{k}=0.44\), \(g = 9.8\space m/s^{2}\)

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Answer:

a) The speed of the snowboarder at the base is approximately \(46.2\space m/s\).
b) The distance the snowboarder travels before stopping is approximately \(248\space m\).