What is the spring constant of a spring that stores 25 J of elastic potential energy when compressed by 7.5 cm from its relaxed length?
I know that I should chnage 7.5 cm into .075 m. Is this equation correct? U(x)= 1/2kx^2. I should have gotten 89N/cm for the correct answer but it isn't working out.
U= 1/2 k x^2
k= 2U/x^2 = 2*25/.075^2=8900N/m^2 This does not convert to the answer you cited.
Or you can work it in cm..
U= 1/2 k x^2
2U/x^2 = 2*25/7.5^2 N/cm
To find the spring constant (k) of a spring, we can use the equation:
U = 1/2 k x^2
Where U is the elastic potential energy stored in the spring, k is the spring constant, and x is the displacement from the relaxed length.
In this case, we are given that the spring stores 25 J of elastic potential energy and is compressed by 7.5 cm (which should indeed be converted to 0.075 m for consistency with SI units).
So, we can substitute these values into the equation:
25 J = 1/2 k (0.075 m)^2
To solve for k, first, square the displacement:
(0.075 m)^2 = 0.005625 m^2
Then, rearrange the equation:
25 J = 1/2 k (0.005625 m^2)
Now, isolate the spring constant:
k = (25 J) / (1/2 * 0.005625 m^2)
Simplifying further:
k = 2 * (25 J) / 0.005625 m^2
k ≈ 8888.89 N/m
Therefore, the spring constant of the spring is approximately 8888.89 N/m.