You throw an 18.0 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 17.0
m/s. Part A Use the work-energy principle to find the rock's speed just as it left the ground. Express your answer in meters per second. vground
1 answer:
Answer:
Vi = 24.14 m/s
Explanation:
If we apply Law of Conservation of Energy or Work-Energy Principle here, we get: (neglecting friction)
Loss in K.E of the Rock = Gain in P.E of the Rock
(1/2)(m)(Vi² - Vf²) = mgh
Vi² - Vf² = 2gh
Vi² = Vf² + 2gh
Vi = √(Vf² + 2gh)
where,
Vi = Rock's Speed as it left the ground = ?
Vf = Final Speed = 17 m/s
g = 9.8 m/s²
h = height of rock = 15 m
Therefore,
Vi = √[(17 m/s)² + 2(9.8 m/s²)(15 m)]
Vi = √583 m²/s²
<u>Vi = 24.14 m/s</u>
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Answer:
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Explanation:
Given:
Coefficient of static friction = 0.25
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Man will slide down if
tan13° > Coefficient of static friction
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