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4 years ago

5

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-hi

gh 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.

1 answer:

Vlada [557]4 years ago

5 0

Answer:

Vmax=11.53 m/s

Explanation:

from conservation of energy

$E_A} =E_{B}$

Spring potential energy =potential energy due to elevation

0.5*k*x²= mg $(h_{B}-h_{A} )$ =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

$E_A} =E_{C}$

spring potential energy=change in kinetic energy

0.5*k"*x²=0.5*m* $V_{max}^{2}$

10812.88 *2.3²=430* $V_{max}^{2}$

$V_{max}$ =11.53 m/s

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