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pav-90 [236]
2 years ago
15

A well is being dug. A 4.5-kg bucket is filled with 28.0 kg of dirt and pulled vertically upward at a constant speed through a d

istance of 9.0 m. Determine the work done on the filled bucket in raising out of the hole.
Physics
1 answer:
andreyandreev [35.5K]2 years ago
4 0

The work done on the filled bucket in raising out of the hole is 2, 925 Joules

<h3>How to determine the work done</h3>

Using the formula:

Work done = force * distance

Note that force = mass * acceleration

F = mg + ma

F = 4. 5 * 10 + 28 * 10

F = 45 + 280

F = 325 Newton

Distance = 9m

Substitute into formula

Work done = 325 * 9

Work done = 2, 925 Joules

Therefore, the work done is 2, 925 Joules

Learn more about work done here:

brainly.com/question/25573309

#SPJ1

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A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
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Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

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Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

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v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

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