Question

Determine a energia cinética associada ao movimento de um homem e sua motocicleta, cuja massa é igual a 450 kg e velocidade igual a 36 km/h.

Answer 1

$\Large\boxed{\begin{array}{l}\sf v=36~km/h~\longrightarrow\!\!\!\!\!\!\!\!\!\!\!\!^{\!\!\div3,6}=10~m/s\\\sf E_c=\dfrac{1}{2}\cdot m\cdot v^2\\\\\sf E_c=\dfrac{1}{2}\cdot 450\cdot10^2\\\\\sf E_c=\dfrac{45000}{2}\\\\\sf E_c=22500~J\end{array}}$

Answer 2

$\Large\displaystyle\text{ \begin{gathered} \hookrightarrow \boxed{ \boxed{\bf{ E_C = 22500 \: J}}} \end{gathered} }$

• Explicação:

$\Large\displaystyle\text{ \begin{gathered} \sf{ E_C = \frac{ m \cdot v {}^{2} }{2} } \end{gathered} }$

Ec ⇒energia cinética (J)

m⇒massa (kg)

v⇒velocidade (m/s)

$\Large\displaystyle\text{ \begin{gathered} \sf{ E_C = \frac{450 \cdot10 {}^{2} }{2} } \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \sf{ E_C = \frac{450 \cdot100}{2} } \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \sf{ E_C = \frac{45000}{2} } \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \boxed{ \boxed{\sf{ E_C = 22500 \: J}}} \end{gathered} }$