A roller coaster car has a mass of 290. kilograms. Starting from rest, the car acquires
3.13 × 10^5 joules of kinetic energy as it descends to the bottom of a hill in 5.3 seconds.
Calculate the speed of the roller coaster car at the bottom of the hill. [Show all work, including the
equation and substitution with units.]

1 Answer

  •      The Definition of Kinetic Energy is given by, we have:

    [tex]E_{c}= \frac{mv^2}{2} [/tex]
         Enteriing the unknowns, we have:

    [tex]E_{c}= \frac{mv^2}{2} \\ 3.13*10^5*J= \frac{290*kg*v^2}{2} \\ v^2= \frac{3.13*10^5*J}{145*kg} \\ v=46.461* \sqrt{ \frac{J}{Kg} } \\ v=46.461* \sqrt{ \frac{N*m}{Kg} } \\ v=46.461* \sqrt{ \frac{ \frac{Kg*m}{s^2} *m}{Kg} } \\ \boxed {v=46.461*m/s}[/tex]

    Obs: approximate results.

    If you notice any mistake in my english, please let me know, because i am not native.