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Vesnalui [34]
2 years ago
9

An 80 kg skier stands at the top of a 40-meter slope. She then skis down the slope. What is her approximate speed at the bottom

of the slope if friction is negligible and her kinetic energy is 31360 J?
Physics
1 answer:
irinina [24]2 years ago
5 0

Thank you for that information.  If we know her mass and kinetic energy, then we don't need any of that other business about the height of the slope, her skiing ability, the friction of the snow, what she uses to wax her skis, the color of her parka, or what she had for lunch at the chalet.  The mass and kinetic energy are enough to answer the question.

Kinetic Energy = (1/2) · (mass) · (speed)²

31,360 J = (1/2) · (80 kg) · (speed)²

speed² = (31,360 J) / (40 kg)

speed² = 784 m²/s²

speed = <em>28 m/s</em>

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Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

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Substituting values, we find

a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2

we also have a tangential acceleration, which is given by

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Substituting values,

a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2

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