{"id":21326,"date":"2022-03-24T15:56:05","date_gmt":"2022-03-24T15:56:05","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?page_id=21326"},"modified":"2022-10-02T19:45:45","modified_gmt":"2022-10-02T18:45:45","slug":"dois-carrinhos","status":"publish","type":"page","link":"https:\/\/www.acasinhadamatematica.pt\/?page_id=21326","title":{"rendered":"Dois carrinhos"},"content":{"rendered":"
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\"\"<\/figure>\n<\/div>\n\n

Sintetizando os dados, temos:<\/p>\n\n\n\n\n\n\n\n\n
Carrinho I<\/strong><\/td>\nCarrinho II<\/strong><\/td>\n<\/tr>\n<\/thead>\n
\\({m_1} = 500\\) g<\/td>\n\\({m_2} = 250\\) g<\/td>\n<\/tr>\n
\\({v_0} = 0\\) m\/s<\/td>\n\\({v_0} = 4\\) m\/s    (\\({v_0} = 14,4\\) km\/h)<\/td>\n<\/tr>\n
\\({v_f} = 2\\) m\/s<\/td>\n\\({v_f} = 0\\) m\/s<\/td>\n<\/tr>\n
\\(\\overrightarrow {{F_1}} \\) \u00e9 constante e \\(\\left| {\\overrightarrow {{F_1}} } \\right| = \\left| {\\overrightarrow {{F_2}} } \\right|\\)<\/td>\n\\(\\overrightarrow {{F_2}} \\) \u00e9 constante e \\(\\left| {\\overrightarrow {{F_2}} } \\right| = \\left| {\\overrightarrow {{F_1}} } \\right|\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n
<\/div>\n\n\n

Designando por \\(a\\) o m\u00f3dulo da acelera\u00e7\u00e3o \\(\\overrightarrow {{a_1}} \\) do carrinho I, e dado que \\(\\overrightarrow {{a_2}} = – 2\\overrightarrow {{a_1}} \\), temos;<\/p>\n\n\n\n\n\n\n
Carrinho I<\/strong><\/td>\nCarrinho II<\/strong><\/td>\n<\/tr>\n<\/thead>\n
\\({v_1} = at\\)<\/td>\n\\({v_2} = 4 – 2at\\)<\/td>\n<\/tr>\n
\\({d_1} = \\frac{1}{2}a{t^2}\\)<\/td>\n\\({d_2} = 4t – \\frac{1}{2} \\times 2a{t^2} = 4t – a{t^2}\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n
<\/div>\n\n\n
Quest\u00e3o 1<\/h5>\n

Confirma-se que o m\u00f3dulo do trabalho realizado pela for\u00e7a \\(\\overrightarrow {{F_2}} \\) \u00e9 o dobro do trabalho realizado pela for\u00e7a \\(\\overrightarrow {{F_1}} \\).<\/p>\n

Com efeito, temos (em J):<\/p>\n

\\[\\begin{array}{*{20}{c}}{\\begin{array}{*{20}{l}}{{W_1}}& = &{{E_{{c_f}}} – {E_{{c_i}}}}\\\\{}& = &{\\frac{1}{2} \\times 0,5 \\times \\left( {{2^2} – {0^2}} \\right)}\\\\{}& = &1\\end{array}}&{}&{}&{}&{\\begin{array}{*{20}{l}}{{W_2}}& = &{{E_{{c_f}}} – {E_{{c_i}}}}\\\\{}& = &{\\frac{1}{2} \\times 0,25 \\times \\left( {{0^2} – {4^2}} \\right)}\\\\{}& = &{ – 2}\\end{array}}\\end{array}\\]<\/p>\n\n\n

<\/div>\n\n\n
Quest\u00e3o 2<\/h5>\n

N\u00e3o se confirma que, ap\u00f3s um mesmo deslocamento<\/span>, o carrinho I adquire a velocidade de m\u00f3dulo \\(2\\) m\/s e o carrinho II para.<\/p>\n

Com efeito, temos:<\/p>\n

    \n
  • o carrinho II para quando \\({v_2} = 0\\) m\/s, isto \u00e9, para \\(at = 2\\);<\/li>\n
  • o carrinho I adquire a velocidade de m\u00f3dulo \\(2\\) m\/s quando \\({v_1} = 2\\) m\/s, isto \u00e9, tamb\u00e9m para \\(at = 2\\).<\/li>\n<\/ul>\n

    Considerando \\(at = 2\\) nas equa\u00e7\u00f5es \\({d_1} = \\frac{1}{2}a{t^2}\\) e \\({d_2} = 4t – a{t^2}\\) acima, temos:<\/p>\n

    \\[\\begin{array}{*{20}{l}}{\\frac{{{d_2}\\left( {at = 2} \\right)}}{{{d_1}\\left( {at = 2} \\right)}}}& = &{\\frac{{4t – 2 \\times t}}{{\\frac{1}{2} \\times 2 \\times t}}}\\\\{}& = &{\\frac{{2t}}{t}}\\\\{}& = &2\\end{array}\\]<\/p>\n

    Portanto, o deslocamento observado no carrinho II \u00e9 duplo<\/span> do deslocamento observado no carrinho I<\/strong>.<\/p>\n

    <\/p>\n

    A rela\u00e7\u00e3o entre as quatro vari\u00e1veis \\({v_1}\\), \\({v_2}\\), \\({d_1}\\) e \\({d_2}\\), pode ser explorada na anima\u00e7\u00e3o seguinte, em fun\u00e7\u00e3o do par\u00e2metro  \\(a\\), o m\u00f3dulo da acelera\u00e7\u00e3o \\(\\overrightarrow {{a_1}} \\) do carrinho I.<\/p>\n\n\n

    <\/div>\n\n\n\n