Question

При удлинении пружины на 0,12 см сила упругости составила 6Н. При каком удлинении пружины сила упругости составляет 10Н?​

Answer 1

Дано:
Дельта L=0,12
F=6
Найти:
k=? если F=10
Решение:
F=k*дельта L
6=k*0,12
k=6÷0,12
k=50
10=50*дельта L
Дельта L=10÷50
Дельта L=0,2
Ответ:Дельта L=0,2см
(дельта L и удлинение пружины-это одно и тоже)

Answer 2

Calculating Spring Force

To calculate the elongation of the spring when the force is 10N, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the amount it is stretched or compressed. The formula for Hooke's Law is:

F = kx

Where: - F is the force applied to the spring, - k is the spring constant, and - x is the displacement from the spring's equilibrium position.

Given that the force is 6N when the spring is elongated by 0.12cm, we can use this information to find the spring constant k.

Finding the Spring Constant (k)

Using the given information, we can rearrange Hooke's Law to solve for the spring constant k:

k = F / x

Substitute the given values: - F = 6N - x = 0.12cm = 0.0012m (converting to meters for consistency)

k = 6N / 0.0012m = 5000 N/m.

Calculating Elongation for 10N Force

Now that we have the spring constant k, we can use it to calculate the elongation of the spring when the force is 10N.

F = kx

Rearranging for x: x = F / k

Substitute the given values: - F = 10N - k = 5000 N/m

x = 10N / 5000 N/m = 0.002m = 2mm

So, when the force applied to the spring is 10N, the elongation of the spring will be 2mm.

I hope this helps! If you have further questions, feel free to ask.