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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..

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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.

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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.

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

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a)

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

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\\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}

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

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b)

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

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\\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}

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

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c)

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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.

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

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\\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}

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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 angle of the slope.

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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

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$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

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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.

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De su respuesta en $m$.

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