what is the force constant, k, for this spring?

Question

the spring in a clothes pin is compressed 0.40 cm, storing energy in the form of spring potential evergy. if the spring is compressed twice as far, the spring potential energy increases by 0.0046 J. what is the force constant, k, for this spring?


this would really help if u could show the work. ....please!!!!???!!!

190 m/N is the answer the book gives...i just dont know how to get that answer :(

please keep answering. neither of these two answers actually gave the correct answer. thanks so much

Answer 1

The potential energy stored in a compressed spring is ½kx²
where
x is the displacement from its equilibrium point
k is the spring constant.

First compression be 0.4 cm = 0.004 m
Thus second compression = 2*0.4 cm = 0.008 m

Energy difference = 0.0046 J

½k*0.008² - ½k*0.004² = 0.0046
½k(0.000064 - 0.000016) = 0.0046
½k(0.000048) = 0.0046
k = 2*46/0.48 = 191.667 N/m

Answer 2

EDITED RESPONSE

The problem is what is meant by "twice as far"? That could mean halve the distance which gives ∆s = 0.2cm, or it could mean another 0.4 cm. It is also not clear whether the length given is the distance from uncompressed length or the compressed length itself.
The energy stored in a (linear) spring is E = 0.5*k*s^2; ∆E is the change in energy, ∆s is the change in compression. In this instance, ∆s is one half of 0.40 = 0.20 cm. The energy change is 0.0046J. From the above equation

k = 2*∆E/(s1^2 - s2^2)

The assumption that the length indicated is distance from uncompressed length is the only one that comes close to your answer.

Answer 3

0.5 *k* (8^2 - 4^2) / (1000) ^2 = 0.0046

0.5 *k* (48) / (1000) ^2 = 0.0046

k (24) = 4600

k = 191.7 N/m

The book answer is approximate answer. The above two answers are correct.

To get the exact answer as in the book you must change the energy spent as 0.00456.J

If you use 0.00456

k = 4560/ 24 = 190 N/m

Answer 4

I agree with rusi911.
I think your book has given a rounded answer.
191 is very much closer to 190.

Answer 5

E= 0.5*k*l*l

E1 = 0.5*k * (0.4*(1/100)) * (0.4*(1/100)) -------(1)
E2 = 0.5*k * (0.8*(1/100)) * (0.8*(1/100)) -------(2)

E2-E1 = 0.0046J --------(3)

(2)/(1)
answer is E2 = 4 * E1

so replace E2 in the equation (3) with 4 * E1and you will get E1=0.001534J.

Replace E1 in the equation (1) with the answer.

The final answer is k = 191.75 or somewhere around that.
I'm not 100% sure.