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You are asked to design spring bumpers for the walls of a parking garage. a freely rolling 1200 kg car moving at 0.70 m/s is to compress the spring no more than 7.2×10−2 m before stopping. what should be the force constant of the spring? assume that the spring has negligible mass.

Respuesta :

skyluke89
We can solve the problem by using law of conservation of energy.

The kinetic energy of the car when it is in motion is
[tex]K= \frac{1}{2}mv^2 [/tex]
where m is the car mass and v its velocity.

Then the car hits the spring and it compresses it by a certain [tex]\Delta x[/tex], so all the kinetic energy of the car is converted into elastic potential energy of the spring:
[tex]U= \frac{1}{2}k (\Delta x)^2 [/tex]
where k is the spring constant.

For the conservation of energy, K=U. Also, the problem says that the maximum allowed value for the compression is [tex]\Delta x= 7.2 \cdot 10^{-2} m[/tex]: if we use this value, we can write
[tex] \frac{1}{2}mv^2 = \frac{1}{2}k (\Delta x)^2 [/tex]
that we can solve to find the spring constant k:
[tex]k= \frac{mv^2}{(\Delta x)^2}= \frac{(1200 kg)(0.70 m/s)^2}{(7.2 \cdot 10^{-2}m)^2}= 1.13 \cdot 10^5 N/m[/tex]
shubhamchouhanvrVT

The force constant for the spring will be  [tex]k=1.13\times 10^5\frac{N}{m}[/tex]

What will be the spring force constant?

It is given that

Mass m= 1200kg

Distnace =  [tex]\Delta X=7.2\times 10^{-2}m[/tex]

Speed of car   [tex]V=0.70\dfrac{m}{s}[/tex]

The kinetic energy of the car will be

[tex]KE= \dfrac{1}{2} mV^2[/tex]

The energy stored in the spring will be

[tex]U= \dfrac{1}{2} k\Delta X^2[/tex]

Now from the conservation of energy, the energy transferred by the car will go to the energy stored in the spring.

[tex]KE=U[/tex]

[tex]\dfrac{1}{2} mV^2=\dfrac{1}{2} kX^2[/tex]

[tex]k=\dfrac{mV^2}{(\Delta X^2)}[/tex]

By putting the values in the formula

[tex]k=\dfrac{1200\times 0.70}{7.2\times 10^{-2}}[/tex]

[tex]k=1.13\times 10^5\ \dfrac{N}{m}[/tex]

Thus the force constant for the spring will be  [tex]k=1.13\times 10^5\frac{N}{m}[/tex]

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