Question

Важок підвісили до пружини довжина якої збільшилась на 10 см Знайдіть силу пружності що діє на пружину якщо її жорсткість 2000

Answer 1

Ответ: 200Н

Объяснение:

Дано:

х=10см=0,1м

k=2000Н/м

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F-?

Согласно закона Гука:

F=kx=2000*0,1=200Н

Answer 2

Calculating the Force of Elasticity on the Spring

To calculate the force of elasticity on the spring, 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 exerted by the spring (in newtons) - k is the spring constant or stiffness of the spring (in newtons per meter) - x is the displacement of the spring from its equilibrium position (in meters)

In this case, the length of the spring increased by 10 cm, which is 0.1 meters. The spring constant is given as 2000 N/m.

Calculating the Force of Elasticity

Using the formula F = kx, we can calculate the force of elasticity on the spring:

F = 2000 N/m * 0.1 m

F = 200 N

So, the force of elasticity acting on the spring is 200 newtons.