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applesjgtl

  • 3 years ago

A 1200 kg racecar is driven along a frictionless horizontal surface at a speed of 65 km/hr. If a horizontal cord brings the car to rest in a distance of 2.2 m, what is the elasticity constant (k) of this spring?

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  1. Shane_B
    • 3 years ago
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    Do you have any idea where to start?

  2. applesjgtl
    • 3 years ago
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    Not really.

  3. applesjgtl
    • 3 years ago
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    65km/hr=18.06m/s

  4. Shane_B
    • 3 years ago
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    I'd start by converting the velocity to m/s. Then calculate the KE of the racecar using:\[KE=\frac{1}{2}mv^2\]

  5. Shane_B
    • 3 years ago
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    Since the racecar comes to a rest you know that the change in KE will be the amount of work done on the car and: \[W=\frac{1}{2}kx^2\]

  6. applesjgtl
    • 3 years ago
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    1/2(1200kg)(18.06m/s)^2=195698.16J

  7. Shane_B
    • 3 years ago
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    So:\[KE=\frac{1}{2}(1200kg)(18.06m/s)^2=195698J=Work\space done\]\[195698J=\frac{1}{2}k(2.2m^2)\]Now just solve for k.

  8. Shane_B
    • 3 years ago
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    Small correction there...it should have been:\[195698J=\frac{1}{2}k(2.2m)^2\]

  9. applesjgtl
    • 3 years ago
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    Oh, because it's not the unit that's being squared.

  10. Shane_B
    • 3 years ago
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    Yep

  11. Shane_B
    • 3 years ago
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    Looking at that last equation, what do you get for k?

  12. applesjgtl
    • 3 years ago
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    k=80867.01

  13. applesjgtl
    • 3 years ago
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    What's that unit?

  14. Shane_B
    • 3 years ago
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    It's a constant...no unit. You also have to consider significant figures in your calcs but that's how you solve it.

  15. applesjgtl
    • 3 years ago
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    Not overly concerned with significant figures in this case, but you make a good point. Thanks!

  16. Shane_B
    • 3 years ago
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    no problem :)

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