Question

A 1100kg car moving on a horizontal surface has speed v = 90km/h when it strikes a horizontal colied spring and is brought to rest at a distance of 2.4m. What is the spring stiffness constant of the spring?

Answer 1

Alright, I think I know how to do this problem, but for some reason I keep getting the wrong answer:

Using hooke's law:

\[(1100kg)(9.80m/s^2)/2.4m = 4491N/m\]

I don't think I would need to use velocity in this problem so I didn't include it.

Answer 2

Yes, your calculation of acceleration is wrong
The car goes from 90km/h(25m/s) to 0 in 2.4 m
so, a=(v^2 -u^2)/2s
=130.2m/s^2
now you can use hooke's law to calculate K
K=1100*130.2/2.4
=59675N/m

Answer 3

okay. Thanks for pointing out my mistake.

Answer 4

^ Its saying that its wrong. what does u stand for?

Answer 5

what do you mean?

Answer 6

try conservation of energy
E(initial)=E(final)
1/2*M*V^2=1/2*K*Δl^2
k=m V^2/Δl^2
K=1100*25^2/2.4^2
K=119357.6389 N/m

Answer 7

some thin else too:

Using hooke's law:
(1100kg)(9.80m/s2)/2.4m=4491N/m THAT IS NOT HOOKE LOW

K=1100*130.2/2.4 =59675N/m THAT IS NOT NOOKE LOW

HOOK: F(from spring)= K * Δl(diferance of leght of spring)

Answer 8

^^^ rahathsbhat I'm doing my homework on as website, so if the answer is wrong it'll let me know.

I'll try your method mos and see if it works. Thanks for the clarification regarding Hooke's Law.

Answer 9

^^Its correct. Thank you mos!