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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.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "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..
\nLa aceleracion de gravedad es $g = 9.8 \\mathrm{ms^{-2}}$.
", "advice": "Se puede dibujar un diagrama para mostrar las fuerzas que actúan sobre la partícula.
\n\nAquí 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.
\nLa pendiente es áspera por lo que la fricción ($\\mu N $ N) actúa cuesta abajo, en contra de la dirección del movimiento.
\nLa 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}
La fuerza de reacción normal entre la partícula y el plano es $\\var{precround(R,3)} \\mathrm{N}$.
\nPara 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}
Por lo tanto, la desaceleración de la partícula es $\\var{precround(-a,3)} \\mathrm{ms^{-2}}$ ya que es el negativo de la aceleración.
\nLa 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.
\nSabemos $u = \\var{u}, v= 0$ 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}
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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", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "a<0", "maxRuns": 100}, "ungrouped_variables": ["g", "mu", "theta", "u", "mass", "R", "a", "s"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "¿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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\nDe su respuesta en unidades de $\\mathrm{ms^{-2}}$.
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\nDe su respuesta en $m$.
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