## anonymous 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?

1. Shane_B

Do you have any idea where to start?

2. anonymous

Not really.

3. anonymous

65km/hr=18.06m/s

4. Shane_B

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

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. anonymous

1/2(1200kg)(18.06m/s)^2=195698.16J

7. Shane_B

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

Small correction there...it should have been:$195698J=\frac{1}{2}k(2.2m)^2$

9. anonymous

Oh, because it's not the unit that's being squared.

10. Shane_B

Yep

11. Shane_B

Looking at that last equation, what do you get for k?

12. anonymous

k=80867.01

13. anonymous

What's that unit?

14. Shane_B

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

15. anonymous

Not overly concerned with significant figures in this case, but you make a good point. Thanks!

16. Shane_B

no problem :)