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A 2.6 kg rock is dropped from a height of 10 m. With what speed will it strike the ground. Ignore air resistance. Solve using conservation of energy (start with Ebefore = Eafter).​

User Christian Dahlqvist
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1 Answer

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12 votes

Final answer:

To solve using the Principle of Conservation of Energy, equate the initial potential energy of the rock to the final kinetic energy when it strikes the ground. By plugging in the given values and simplifying, we find that the rock will strike the ground with a speed of approximately 14 m/s.

Step-by-step explanation:

To solve this problem using the Principle of Conservation of Energy, we equate the initial potential energy of the rock to the final kinetic energy when it strikes the ground.

Initial potential energy (Ebefore) = Final kinetic energy (Eafter)

mgh = 1/2 mv2,

where m is the mass of the rock, g is the acceleration due to gravity, h is the height the rock is dropped from, and v is the final velocity of the rock when it strikes the ground.

Plugging in the given values:

2.6 kg * 9.8 m/s2 * 10 m = 1/2 * 2.6 kg * v2,

Simplifying the equation gives:

254.8 joules = 1.3 * v2,

Divide both sides by 1.3 kg to solve for v2:

v2 = 254.8 joules / 1.3 kg,

v2 ≈ 196 m2/s2,

Taking the square root of both sides to solve for v:

v ≈ √(196 m2/s2) = 14 m/s.

The rock will strike the ground with a speed of approximately 14 m/s.

User Basin
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