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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?

Physics
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Do you have any idea where to start?
Not really.
65km/hr=18.06m/s

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Other answers:

I'd start by converting the velocity to m/s. Then calculate the KE of the racecar using:\[KE=\frac{1}{2}mv^2\]
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: \[W=\frac{1}{2}kx^2\]
1/2(1200kg)(18.06m/s)^2=195698.16J
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.
Small correction there...it should have been:\[195698J=\frac{1}{2}k(2.2m)^2\]
Oh, because it's not the unit that's being squared.
Yep
Looking at that last equation, what do you get for k?
k=80867.01
What's that unit?
It's a constant...no unit. You also have to consider significant figures in your calcs but that's how you solve it.
Not overly concerned with significant figures in this case, but you make a good point. Thanks!
no problem :)

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