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Irina18 [472]
3 years ago
15

Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work? Move the machine to a

different gravitational field. Increase the friction between its moving parts. Reduce the friction between its moving parts. Redefine the machine’s system boundaries.
Physics
2 answers:
Lady bird [3.3K]3 years ago
7 0
Reduce the friction. Since the total energy is conserved, the only way to improve its work capacity is by reducing energy that doesnt go into work.
Ulleksa [173]3 years ago
4 0

Reduce the friction between its moving parts

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kramer

Part A)

As we know that spring force is given by

F = kx

here x = stretch in the spring from natural length

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so acceleration = 0

Part b)

When spring is compressed from its natural length it will have elastic potential energy in it

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U = \frac{1}{2}kx^2

now we know that there is no friction in it so maximum kinetic energy of the launcher must be equal to the elastic potential energy of the spring

KE = \frac{1}{2}kx^2

here we have

k = 70 N/m

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KE = \frac{1}{2}(70)(0.4)^2

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Part c)

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KE = \frac{1}{2} mv^2

5.6 = \frac{1}{2}0.3v^2

v = 6.11 m/s

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3 years ago
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Answer: The loss of energy due to friction is equal to 1,253 J.

Explanation:

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After coming to a  flat landing, his final speed is 7.8 m/s, so the final kinetic energy is as follows:

K₂ = 1/2 m v₂² = 1/2. 58.0 Kg. (7.8)² (m/s)² = 1,764 J

Now, when skying down the slope the increase in kinetic energy only can come from another type of energy, in this case, gravitational potential energy.

If we take the ground flat level as a Zero reference, the initial gravitational potential energy, can be written as follows, by definition:

U₁ = m.g. h (1)

Now, we don't know the value of the height h, but we know that the incline has a 18º angle above the horizontal, and that the distance travelled along the incline is 15 m.

By definition, the sinus of an angle, is equal to the proportion between the height and the hypotenuse , so we can write the following equation:

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Replacing in (1), we get:

U₁ = 58.0 Kg. 9.8 m/s². 4.6 m = 2,641 J

So, we can get the total initial mechanical energy, as follows:

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After arriving to the flat zone, all potential energy has become in kinetic energy, even though not completely, due to the effect of friction.

This remaining kinetic energy can be written as follows:

E₂ = K₂ = 1,764 J

The difference E₂-E₁, is the loss of energy due to friction forces acting during the travel along the 15 m path, and is as follows:

ΔE= E₂ - E₁ = 1,764 J - 3,017 J = -1,253 J

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jolli1 [7]

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