Question

In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it starts to careen down the steeply sloped track. In one event, the sled reaches a top speed of 9.2 m/s before starting down the initial part of the track, which is sloped downward at an angle of 9.0 ∘.

 Part A

 What is the sled's speed after it has traveled the first 140 m?

Answer 1

Gravitational acceleration \(=\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\)

Mass of the sled \(=\mathrm{m}\)

Speed of the sled before starting down the initial part of the track \(=\mathrm{v}_{\mathrm{i}}=9.2 \mathrm{~m} / \mathrm{s}\)

Angle of the slope \(=\theta=9^{\circ}\)

Speed of the sled after it has travelled the first \(140 \mathrm{~m}=\mathrm{v}_{\mathrm{f}}\)

Distance the sled has travelled \(=\mathrm{L}=140 \mathrm{~m}\)

Height through which the sled has fallen \(=\mathrm{h}\)

\(\mathrm{h}=\mathrm{L} \operatorname{Sin} \theta\)

\(h=(140) \operatorname{Sin}(9)\)

\(\mathrm{h}=21.9 \mathrm{~m}\)

By conservation of energy the initial potential and kinetic energy of the sled is equal to the final kinetic energy of the sled.

\(m g h+\frac{m v_{i}^{2}}{2}=\frac{m v_{f}^{2}}{2}\)

\(g h+\frac{v_{i}^{2}}{2}=\frac{v_{f}^{2}}{2}\)

\(2 g h+v_{i}^{2}=v_{f}^{2}\)

\(v_{f}=\sqrt{2 g h+v_{i}^{2}}\)

\(v_{f}=\sqrt{2(9.8)(21.9)+(9.2)^{2}}\)

\(\mathrm{v}_{\mathrm{f}}=22.7 \mathrm{~m} / \mathrm{s}\)

A) Speed of the sled after it has travelled the first \(140 \mathrm{~m}=22.7 \mathrm{~m} / \mathrm{s}\)