Question

Условие задания: 26. Под действием силы давления вагона 65 кН буферные пружины между вагонами сжимаются на 2 см. Определи силу, с которой давит вагон, если пружины сжались на 5 см,

Answer 1

Ответ: 162,5кН

Объяснение:

Дано:

F=65кН=65000Н

х1=2см=0,02м

х2=5см=0,05м

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F2-?

В виде уравнения закон Гука записывается в следующей форме: F = –kx где F — сила упругого сопротивления струны, x — линейное растяжение или сжатие, а k — так называемый коэффициент упругости.

отсюда k=F/х1=65000/0,02=3250000Н/м

F2=k*х2=3250000*0,05=162500Н=162,5кН

Answer 2

Calculation of Force Exerted by the Wagon

To determine the force exerted by the wagon, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

Hooke's Law can be expressed as:

F = k * x

Where: - F is the force exerted by the spring, - k is the spring constant, and - x is the displacement of the spring.

In this case, the displacement of the buffer springs is given as 2 cm when the force is 65 kN. We can use this information to calculate the spring constant.

Let's assume that the spring constant is denoted by k1.

65 kN = k1 * 2 cm

Converting the displacement to meters:

65 kN = k1 * 0.02 m

Simplifying the equation:

k1 = 65 kN / 0.02 m

k1 = 3250 kN/m

Now, we can use the calculated spring constant to determine the force when the displacement is 5 cm.

Let's assume that the force exerted by the wagon is denoted by F2.

F2 = k1 * 5 cm

Converting the displacement to meters:

F2 = 3250 kN/m * 0.05 m

Simplifying the equation:

F2 = 162.5 kN

Therefore, the force with which the wagon is pressing is 162.5 kN.

Please note that the calculations above are based on the assumption that the behavior of the buffer springs follows Hooke's Law.