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CaHeK987 [17]
4 years ago
15

In one experiment, the students allow the block to oscillate after stretching the spring a distance A. If the potential energy s

tored in the spring is U0, then what is the change in kinetic energy of the block after it is released from rest and has traveled a distance of A2?
Physics
2 answers:
atroni [7]4 years ago
3 0

Answer:

    K = m g (A - A2)

Explanation:

In a block spring system the total energy is the sum of the potential energy plus the kinetic energy, for maximum elongation all the energy is potential

         Em = U₀ = m g A

For when the system is at an ele

Elongation A2 less than A, energy has two parts

        Em = K + U₂

       K = Em –U₂

We substitute

     K = m g A - m gA2

    K = m g (A - A2)

victus00 [196]4 years ago
3 0

Answer:

ΔK.E = U0

Explanation:

Solution:-

- The spring constant = k

- The amount of distance displaced initially xo = A

- The potential energy stored at point xo = U0

- Considering the mass-spring system in isolation there are no external forces like viscous drag or friction acting, then we can safely apply the principle of conservation of energy.

- Which states:

                                       ΔK.E = ΔP.E

Where,  ΔK.E: The change in kinetic energy

             ΔP.E: The change in potential energy

                                      ΔK.E = U0 - Uf

Where,  Uf : Final potential energy.

- The potential energy stored in the spring is given by:

                                     U = 1/2*k*x^2

Where,    x: The displacement of spring from mean position.

- Once the block has been released from displacement x = A/2 about mean position the block travels back to its mean position with displacement x = 0. So the final potential energy when the block has travelled a distance of A/2 is:

                                    Uf = 1/2*k*0^2 = 0

So,

                                   ΔK.E = U0 - Uf

                                   ΔK.E = U0

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A)     E = 145.6 N / C , B)  y= 2,8 10-7 m with a downward direction

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A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

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We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

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          y = y₀ + v_{oy} t + ½ a t²

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          y = ½ a t²

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        E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

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         E = 2  9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

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Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

          y = ½ e x² / m v₀² E

          y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)²   145.6

          y = 28.4375 10⁻⁸ m

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C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

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           F_{e} = - 1.6 10⁻¹⁹ 145.6

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         F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

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Learn more about time elapses between when the bat emits the sound :

<u>brainly.com/question/16931690</u>

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Correction question:

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