Question

При нагрузке в 200Н пружина динамометра удлинилась на 0,5см. На сколько удлинилась пружина при нагрузке в 700Н?

Answer 1

Дважды записываем закон Гука (для силы в 200Н и для силы в 700 Н соответственно)
$\left \{ {{F_1=k\cdot \Delta l_1} \atop {F_2=k\cdot \Delta l_2}} \right.$
Решая систему относительно второго удлинения, находим:
$\Delta l_2=\Delta l_1\cdot \frac {F_2}{F_1}=1,75$ (cм).
Это ответ.

Answer 2

Calculation of Spring Extension

To calculate the extension of a spring under different loads, we can use Hooke's Law, which states that the extension of a spring is directly proportional to the applied force. The formula for Hooke's Law is:

F = k * x

Where: - F is the applied force (in Newtons) - k is the spring constant (in N/m) - x is the extension of the spring (in meters)

In this case, we are given that the spring extended by 0.5 cm (or 0.005 m) when a load of 200 N was applied. We need to find the extension of the spring when a load of 700 N is applied.

To find the spring constant (k), we can rearrange the formula as follows:

k = F / x

Using the given values, we can calculate the spring constant:

k = 200 N / 0.005 m = 40000 N/m

Now, we can use the spring constant to find the extension of the spring when a load of 700 N is applied:

F = 700 N

k = 40000 N/m

x = ?

Using Hooke's Law, we can rearrange the formula to solve for x:

x = F / k

Substituting the given values:

x = 700 N / 40000 N/m = 0.0175 m

Therefore, the spring will extend by 0.0175 meters (or 1.75 cm) when a load of 700 N is applied.