## applesjgtl Group Title 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? one year ago one year ago

1. Shane_B Group Title

Do you have any idea where to start?

2. applesjgtl Group Title

Not really.

3. applesjgtl Group Title

65km/hr=18.06m/s

4. Shane_B Group Title

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 Group Title

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 Group Title

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

7. Shane_B Group Title

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 Group Title

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

9. applesjgtl Group Title

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

10. Shane_B Group Title

Yep

11. Shane_B Group Title

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

12. applesjgtl Group Title

k=80867.01

13. applesjgtl Group Title

What's that unit?

14. Shane_B Group Title

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 Group Title

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

16. Shane_B Group Title

no problem :)