Un bloque de 3 kg en reposo se deja libre a una altura de 5 m sobre una rampa curva y sin rozamiento. Al pie de la rampa se encuentra un resorte de constante k = 400 N/m, como se muestra en la fig. El objeto desliza por la rampa y llega a chocar contra el resorte comprimiéndolo una distancia x antes de que quede en reposo momentáneamente. Determinar: a) La velocidad con la que el bloque alcanza al resorte. ____________________ b) La distancia x que el bloque comprime al resorte. __________________ c) La velocidad con la que el bloque es expulsado por el resorte. ____________________ d) La altura que alcanza sobre la parte curva. ________________ e) ¿Alcanzará la misma altura si la rampa no está libre de rozamiento? ___________________

Answers

Answer 1

Answer:

a) La velocidad del bloque cuando llega al resorte es de aproximadamente 9,9 m / s

b) La distancia a la que se comprime el resorte es de aproximadamente 0,86 m

c) La velocidad con la que el resorte expulsa el bloque es de aproximadamente 9,9 m / s

d) La altura que alcanza el bloque es de 5 metros.

e) El bloque no alcanzará la misma altura si la rampa no está libre de fricción

Explanation:

a) Los parámetros dados del bloque son;

La masa del bloque, m = 3 kg

La altura a la que se coloca el bloque, h = 5 m

La constante de resorte, k = 400 N / m

La aceleración debida a la gravedad, g = 9,8 m / s²

La energía potencial de un cuerpo, P.E. = m · g · h

Por tanto, la energía potencial inicial del bloque, P.E. se da como sigue;

P.E. = 3 kg × 9,8 m / s² × 5 m = 147 julios

P.E. = 147 julios

La energía cinética del bloque al pie de la rampa, K.E. = 1/2 · m · v²

Dónde;

v = La velocidad del bloque cuando llega al resorte

Por lo tanto, para el bloque dado tenemos;

K.E. = 1/2 · m · v² = 1/2 × 3 kg × v²

Por el principio de conservación de la energía, tenemos;

El PE. del bloque en reposo a una altura de 5 m = La energía cinética al pie de la rampa. K.E.

∴ P.E. = K.E.

147 J = 1/2 × 3 kg × v²

v² = 147 J / (1/2 × 3 kg) = 98 m² / s²

v = √ (98 m² / s²) = 7 · √2 m / s

v = 7 · √2 m / s ≈ 9,9 m / s

b) La energía recibida por el resorte comprimido, E = 1/2 · k · x²

Dónde;

k = La constante del resorte = 400 N / m

x = La distancia a la que se comprime el resorte

Por el principio de conservación de la energía, tenemos;

La energía recibida por el resorte comprimido, E = La energía potencial inicial del resorte, P.E.

∴ E = 1/2 · k · x² = P.E.

De lo que tenemos;

E = 1/2 × 400 N / m × x² = 147 julios

x² = 147 Julios / (1/2 × 400 N / m) = 0,735 m²

x = √ (0,735 m²) = 0,7 · √ (3/2) m ≈ 0,86 m

La distancia a la que se comprime el resorte = x ≈ 0.86 m

c) La velocidad con la que el resorte expulsa el bloque se indica a continuación;

La energía en el resorte = 1/2 · k · x² = La energía cinética dada al bloque, 1/2 · m · v²

∴ 1/2 · k · x² = 1/2 · m · v²

∴ La velocidad con la que el bloque es expulsado por el resorte, v = La velocidad con la que el bloque llega al resorte = 7 · √2 m / s

La velocidad con la que el resorte expulsa el bloque, v = 7 · √2 m / s ≈ 9,9 m / s

d) La altura que alcanza el bloque también viene dada por la siguiente relación anterior;

P.E. = K.E.

∴ m · g · h = 1/2 · m · v²

v = 7 · √2 m / s

De donde tenemos h = La altura inicial del bloque en la rampa = 5 metros

e) El bloque no alcanzará la misma altura si la rampa no está libre de fricción porque se utilizará energía para superar la fuerza de fricción

Answer 2

a) La velocidad final del bloque es aproximadamente 9.903 metros por segundo.

b) El resorte se deforma 0.858 metros.

c) Por el principio de la conservación de energía y sabiendo la ausencia de fuerzas disipativas, la velocidad del objeto expulsado del resorte es aproximadamente 9.903 metros por segundo.

d) Por el principio de la conservación de energía y si existieran fuerzas disipativas, la altura máxima sería menor a la hallada en el punto a).

a) Conforme a la situación de este problema, la energía cinética traslacional final ([tex]K[/tex]), en joules, es igual a la energía potencial gravitacional inicial ([tex]U[/tex]), en joules.

[tex]U = K[/tex] (1)

Por las definiciones de las energías cinética traslacional y potencial gravitacional expandimos la ecuación anterior:

[tex]m\cdot g\cdot h = \frac{1}{2}\cdot m\cdot v^{2}[/tex] (1)

Ahora despejamos la velocidad de esa ecuación:

[tex]v = \sqrt{2\cdot g\cdot h}[/tex]

Donde:

[tex]m[/tex] - Masa del bloque, en kilogramos.[tex]g[/tex] - Aceleración gravitacional, en metros por segundo al cuadrado.[tex]h[/tex] - Altura inicial del bloque, en metros.[tex]v[/tex] - Velocidad final del bloque, en metros por segundo.

Si sabemos que [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] y [tex]h = 5\,m[/tex], entonces la velocidad final del bloque es:

[tex]v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}[/tex]

[tex]v\approx 9.903\,\frac{m}{s}[/tex]

La velocidad final del bloque es aproximadamente 9.903 metros por segundo.

b) Por el principio de conservación de la energía, la energía cinética traslacional inicial es igual a la energía potencial elástica final, cuyas fórmula es la siguiente:

[tex]\frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)

Where:

[tex]k[/tex] - Constante de resorte, en newtons por metro.[tex]x[/tex] - Deformación del resorte, en metros.

Ahora despejamos la deformación del resorte:

[tex]x = \sqrt{\frac{m}{k} }\cdot v[/tex] (3)

Si sabemos con [tex]k = 400\,\frac{N}{m}[/tex], [tex]m = 3\,kg[/tex] y [tex]v \approx 9.903\,\frac{m}{s}[/tex], entonces la deformación del resorte es:

[tex]x = \sqrt{\frac{3\,kg }{400\,\frac{N}{m} } }\cdot \left(9.903\,\frac{m}{s} \right)[/tex]

[tex]x \approx 0.858\,m[/tex]

El resorte se deforma 0.858 metros.

c) Por el principio de la conservación de energía y sabiendo la ausencia de fuerzas disipativas, la velocidad del objeto expulsado del resorte es aproximadamente 9.903 metros por segundo.

d) Por el principio de la conservación de energía y si existieran fuerzas disipativas, la altura máxima sería menor a la hallada en el punto a).

Invitamos cordialmente a ver este problema sobre el principio de conservación de la energía: https://brainly.com/question/16582988


Related Questions

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Answers

Answer:

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Explanation:

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Answers

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Mathematically, Ohm's law is given by the formula;

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Answers

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Consider the north direction as positive and the subscript 1 will be used for car 1 and the subscript 2 for the second car

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Final moment. Right after the crash

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let's calculate

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          v₀₂ = -2.67 m / s

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Answers

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Answers

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The original volume of the sample of hydrogen is 250 mL. The correct option is the fourth option 250mL

From the question,

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