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You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12-m-high hill, then descends 18 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.3 m and that a loaded car will have a maximum mass of 430 kg . For safety reasons, the spring constant should be 13 % larger than the minimum needed for the car to just make it over the top.
a) what spring constant (k) should you specify ?b) What is the maximum speed of a 350 kg car if the spring iscompressed the full amount.

Respuesta :

Hania12

Answer:

Vmax=11.53 m/s

Explanation:

from conservation of energy

      [tex]E_A} =E_{B}[/tex]

     Spring potential energy =potential energy due to elevation

   0.5*k*x²= mg[tex](h_{B}-h_{A} )[/tex]=mgh

   0.5*k*2.3²= 430*9.81*6

         k=9568.92 N/m

For safety reason

                                 k"=1.13 *k= 1.13*9568.92

                                    k"=10812.88 N/m

agsin from conservation of energy

      [tex]E_A} =E_{C}[/tex]

    spring potential energy=change in kinetic energy

   0.5*k"*x²=0.5*m*[tex]V_{max}^{2}[/tex]

      10812.88 *2.3²=430*[tex]V_{max}^{2}[/tex]

           [tex]V_{max}[/tex]=11.53 m/s

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