// Numbas version: finer_feedback_settings {"name": "FIS5047_S4_Numbas", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Energ\u00eda", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", ""], "variable_overrides": [[], []], "questions": [{"name": "Conservacion Energ\u00eda", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ivan Munoz", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18334/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "

Un acróbata sale del extremo de una rampa en una motocicleta con
una rapidez de {velocidad} m/s. Si su rapidez es de {velocidadmax} m/s cuando alcanza el punto más alto de la
trayectoria,Desprecie la fricción y la resistencia del aire.

", "advice": "

Supongamos que la rampa tiene una altura $h$ por lo tanto la energía con la que sale la moto será,

\n

$E_i=\\frac{1}{2}mv_i^2+mgh$

\n

\n

Al llegar a su punto mas alto, la moto tendra energía,

\n

$E_f=\\frac{1}{2} mv_f^2+m g (h+d)$

\n

donde $d$ es la altura que llega a tener la moto. Debido a que no hay roce, hay conservacion de energía. Por lo tanto,

\n

\n

$E_ii=E_f\\Rightarrow \\frac{1}{2}mv_i^2+mgh=\\frac{1}{2} mv_f^2+m g h+mgd\\Rightarrow d=\\frac{v_i^2-v_f^2}{2}=\\var{resp}$

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¿cuál es la altura máxima que alcanza?

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Un esquiador inicia desde el reposo en la parte superior de un plano inclinado de {altura}m de alto. En la parte inferior del plano, el
esquiador encuentra una superficie horizontal donde el coeficiente de fricción cinética entre los esquíes y la nieve es {roce}.

\n

", "advice": "

a)  La energía en la parte superior será $E_i==mgh=m *\\var{gravity}*\\var{altura}$ al llegar al final del plano inclinado, tendra solo energia cinetica

\n

$E_c=\\frac{1}{2 }mv^2$.

\n

Dado que no hay fuerzas no conservativas, la energia se conserva, por lo tanto $mgh=\\frac{1}{2} mv^2$.

\n

Por lo tanto $v=\\sqrt{2gh}=\\var{velocity}$

\n

b) Al desplazarese por la superficie con roce, el cambio de energía cinetica será.

\n

$\\Delta E_c=E_{cf}-E_{ci}=-E_{ci}=-\\frac{1}{2} m v^2$

\n

Dado que hay fuerza no conservativa, el roce. El cambio de energía cinetica es igual al trabajo hecho por el roce,

\n

$W_r=-F_r d=-N \\mu d=-mg d=-\\frac{1}{2}mv^2$

\n

Por lo tanto,

\n

\n

$d=\\frac{v^2}{2g}=\\var{distancia}$

\n

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The mass of the block.

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Determine la rapidez del esquiador en la parte inferior.

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¿Qué distancia recorre el esquiador sobre la superficie horizontal antes de llegar al reposo?

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