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A los estudiantes se les da la ecuación de una parábola y se les pide que identifiquen su gráfica. Se les da la curva parabólica, una exponencial, una hipérbola y una recta a elegir.

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Cual de los siguientes graficos representa $y = \\var{pchoicedispa} \\text{x}^2  \\var{choosesignb}  \\var{choosesign}$?

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The equation given is for a parabola (quadratic).

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Una partícula se proyecta hacia arriba de un plano rugoso a una velocidad dada. Dado el ángulo de la pendiente y el coeficiente de fricción, encuentre la distancia que recorre la partícula antes de llegar al reposo instantáneo.

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Una particula de masa $\\var{mass} \\mathrm{kg}$ es lanzada con velocidad $\\var{u}\\ \\mathrm{ms^{-1}}$ por un plano inclinado. El coeficiente de roce entre la particula y el roce es $\\var{mu}$. El plano posee una inclinacion $\\theta = \\var{theta}^{\\circ}$ con respecto a la horizonta..

\n

La aceleracion de gravedad es $g = 9.8 \\mathrm{ms^{-2}}$.

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Se puede dibujar un diagrama para mostrar las fuerzas que actúan sobre la partícula.

\n

\n

Aquí la partícula se dibuja en tres posiciones, mostrando su velocidad original cuando se proyecta desde el fondo de la pendiente, su posición en algún punto de la pendiente y su posición cuando se detiene instantáneamente más arriba de la pendiente.

\n

La pendiente es áspera por lo que la fricción ($\\mu N $ N) actúa cuesta abajo, en contra de la dirección del movimiento.

\n

a)

\n

La reacción normal $N$, se encuentra resolviendo las fuerzas perpendiculares al plano.

\n

\\begin{align}
R - mg \\cos \\theta & = 0, \\\\
R & = mg \\cos \\theta, \\\\
& = (\\var{mass} \\times 9.8) \\cos (\\var{theta}^{\\circ}), \\\\
&= \\var{precround(R,3)} \\mathrm{N}.
\\end{align}

\n

La fuerza de reacción normal entre la partícula y el plano es $\\var{precround(R,3)} \\mathrm{N}$.

\n

b)

\n

Para encontrar la aceleración de la partícula resolvemos las fuerzas paralelas al plano.

\n

\\begin{align}
- mg \\sin \\theta - \\mu R & = ma, \\\\
- \\var{mass} \\times 9.8 \\sin (\\var{theta}^{\\circ}) - (\\var{precround(R,3)} \\times \\var{mu}) & = \\var{mass}a, \\\\
a & = \\frac{ - \\var{mass} \\times 9.8 \\sin (\\var{theta}^{\\circ}) - (\\var{precround(R,3)} \\times \\var{mu}) }{\\var{mass}}, \\\\
& = \\var{precround(a,3)} \\mathrm{ms^{-2}}.
\\end{align}

\n

Por lo tanto, la desaceleración de la partícula es $\\var{precround(-a,3)} \\mathrm{ms^{-2}}$ ya que es el negativo de la aceleración.

\n

c)

\n

La partícula viajará cuesta arriba hasta que llegue al reposo instantáneo; en este punto su velocidad será $0$. Podemos usar la ecuación $v^2 = u^2 + 2as$ para encontrar la distancia que viajará la partícula.

\n

Sabemos $u = \\var{u}, v= 0$ y $a = \\var{precround(a,3)}$.

\n

\\begin{align}
v^2 & = u^2 + 2as, \\\\
0 & = \\simplify{{u}^2+{precround(2a,3)}s}, \\\\
s & = \\frac{\\var{u}^2}{\\var{precround(-2a,3)}}, \\\\[0.5em]
& = \\var{precround(u^2/(-2*a),3)} \\mathrm{m}.
\\end{align}

\n

La distancia que la partícula viajará por el plano antes de llegar al reposo instantáneo es $\\var{precround(u^2/(-2*a),3)} \\mathrm{m}$.

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The mass of the particle.

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The coefficient of friction between the particle and the plane.

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La aceleracion debido a la gravedad

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The acceleration of the particle along the slope.

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The normal reaction force of the plane against the particle.

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The initial speed of the particle.

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The distance travelled before the particle comes to rest.

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¿Cual es la fuerza normal $N$ entre la partícula y el plano? De su respuesta en Newtons hasta 3 decimales

\n

$N = $ [[0]]

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Usando el valor de $N$ de la parte a), encuentre la aceleracion de la particula a medida que se mueve hacia arriba en el planot

\n

De su respuesta en unidades de $\\mathrm{ms^{-2}}$.

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Encuentre la distancia que se moverá la partícula antes de que llegue al reposo.

\n

De su respuesta en $m$.

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