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?

Question

shane_b · Answer

Do you have any idea where to start?

anonymous · Answer

Not really.

anonymous · Answer

65km/hr=18.06m/s

shane_b · Answer

I'd start by converting the velocity to m/s.  Then calculate the KE of the racecar using:$$KE=\frac{1}{2}mv^2$$

shane_b · Answer

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:  $$W=\frac{1}{2}kx^2$$

anonymous · Answer

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

shane_b · Answer

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.

shane_b · Answer

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

anonymous · Answer

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

shane_b · Answer

Yep

shane_b · Answer

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

anonymous · Answer

k=80867.01

anonymous · Answer

What's that unit?

shane_b · Answer

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

anonymous · Answer

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

shane_b · Answer

no problem :)