Question

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.

Answer 1

To find the spring constant (k) for the roller coaster, calculate the minimum value using the energy conservation equation. Then, add 13% to this minimum value for safety. To find the maximum speed of the car when the spring is fully compressed, use the conservation of mechanical energy equation.

Step-by-step explanation:

To determine the spring constant (k) for the roller coaster, we need to calculate the minimum value first. The minimum value can be found by considering the energy conservation equation at the top of the 12 m hill. At the top, the gravitational potential energy is converted into the potential energy stored in the compressed spring. Therefore, we have:

mgh = 0.5kx^2

where m is the maximum mass of the loaded car, and x is the maximum compression of the spring. Rearranging the equation, we get:

k = 2(mgh)/(x^2)

Now, we can substitute the values given in the question (m = 430 kg, h = 12 m, x = 2.3 m) to find the minimum spring constant. To ensure safety, we need to add 13% to this minimum value. For part 'b', we can use the conservation of mechanical energy to find the maximum speed of the car when the spring is compressed the full amount. Since the spring is fully compressed, all the potential energy stored in the spring will be converted into kinetic energy when the car is released. Therefore, we have:

0.5kx^2 = 0.5mv^2

Rearranging the equation, we get:

v = sqrt((kx^2)/m)

Substituting the values given in the question (m = 350 kg, k = 1.13 times the minimum value found in part 'a', x = 2.3 m), we can calculate the maximum speed of the car.

Answer 2

Vmax=11.53 m/s

Step-by-step 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