Question

um vagão de montanha russa de 300 kg esta inicialmente a 20 km/h quando ira descer um declive de 30 m de altura por 10 m de extensão sabendo que no final da descida, 20% da energia foi dissipada à velocidade aproximada do vagão quando terminada a descida é de:
Dados:g=10 m\Segundo ao quadrado.

Answer 1

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•   massa do vagão:   $\mathsf{m=300~kg;}$

•   velocidade inicial:

$\mathsf{v_i=20~km/h}\\\\ \mathsf{v_i=\dfrac{20~km}{3\,600~s}}\\\\\\ \mathsf{v_i=\dfrac{20\cdot 1000~m}{3\,600~s}}\\\\\\ \mathsf{v_i=\dfrac{20\,000~m}{3\,600~s}}\begin{array}{c}^{\mathsf{\div 400}}\\^\mathsf{\div 400} \end{array}\\\\\\ \mathsf{v_i=\dfrac{50}{9}~m/s\approx 5,\!56~m/s;}$


•   altura inicial:   $\mathsf{h_i=30~m;}$

•   velocidade final:   $\mathsf{v_f;}$

•   altura inicial:   $\mathsf{h_f=0;}$

•   aceleração da gravidade:   $\mathsf{g=10~m/s^2.}$

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•   Energia mecânica inicial

$\mathsf{E_{mi}=mgh_i+\dfrac{1}{2}mv_i^2}\\\\\\ \mathsf{E_{mi}=300\cdot 10\cdot 30+\dfrac{1}{2}\cdot 300\cdot \left(\dfrac{50}{9}\right)^{\!2}}\\\\\\ \mathsf{E_{mi}=90\,000+150\cdot \dfrac{2\,500}{81}}\\\\\\ \mathsf{E_{mi}=90\,000+50\cdot \dfrac{2\,500}{27}}$

$\mathsf{E_{mi}=\dfrac{2\,430\,000+50\cdot 2\,500}{27}}\\\\\\ \mathsf{E_{mi}=\dfrac{2\,430\,000+125\,000}{27}}\\\\\\ \mathsf{E_{mi}=\dfrac{2\,555\,000}{27}}\\\\\\ \mathsf{E_{mi}\approx 94\,629,\!6~J}\qquad\quad\checkmark$


•   Energia mecânica final:

$\mathsf{E_{mf}=mgh_f+\dfrac{1}{2}mv_f^2}\\\\\\ \mathsf{E_{mf}=300\cdot 10\cdot 0+\dfrac{1}{2}\cdot 300\cdot v_f^2}\\\\\\ \mathsf{E_{mf}=150\cdot v_f^2}\qquad\quad\checkmark$


•   Energia dissipada:

$\mathsf{W_{nc}=20\%\cdot E_{mi}}\qquad\quad\checkmark$

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Pela conservação de energia, devemos ter

$\mathsf{E_{mi}=E_{mf}+W_{nc}}\\\\ \mathsf{E_{mi}-W_{nc}=E_{mf}}\\\\ \mathsf{E_{mi}-20\%\cdot E_{mi}=E_{mf}}\\\\ \mathsf{E_{mi}-0,\!20\cdot E_{mi}=E_{mf}}\\\\ \mathsf{0,\!80\cdot E_{mi}=E_{mf}}$

$\mathsf{0,\!80\cdot \dfrac{2\,555\,000}{27}=150\cdot v_f^2}\\\\\\ \mathsf{\dfrac{2\,044\,000}{27}=150\cdot v_f^2}\\\\\\ \mathsf{v_f^2=\dfrac{2\,044\,000}{27}\cdot \dfrac{1}{150}}\\\\\\ \mathsf{v_f^2=\dfrac{2\,044\,000}{4\,050}}$

$\mathsf{v_f=\sqrt{\dfrac{2\,044\,000}{4\,050}}}\\\\\\ \mathsf{v_f\approx \sqrt{504,\!7}}\\\\ \mathsf{v_f\approx 22,5~m/s}\\\\ \mathsf{v_f\approx 22,5\cdot 3,\!6~km/h}\\\\\\ \boxed{\begin{array}{c}\mathsf{v_f\approx 81~km/h} \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a resposta.}$


Bons estudos! :-)