Question

Когда четыре человека массой по 70 кг садятся в автомобиль, пружина амортизатора автомобиля сжимается на 2,5 см Найдите жесткость одной пружины, если всего пружин — четыре

Answer 1

Сила тяжести распределяется равномерно тогда, тогда проекция 2 ого закона ньютона но ось х имеет вид: 4kx=4mg
kx=mg
k=mg/x=10*70/0.025=28000 н/м

Answer 2

Calculation of the Spring Stiffness

To find the stiffness of one spring, 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. Mathematically, this can be expressed as:

F = k * x

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

In this case, when four people with a mass of 70 kg each sit in the car, the spring of the car's shock absorber compresses by 2.5 cm. We can assume that the weight of the people is evenly distributed among the four springs.

To find the stiffness of one spring, we need to calculate the force exerted by the four people on the springs. The force can be calculated using Newton's second law of motion:

F = m * g

Where: - F is the force, - m is the mass, - g is the acceleration due to gravity (approximately 9.8 m/s^2).

Since there are four people, the total mass is 4 * 70 kg = 280 kg. Therefore, the force exerted by the four people is:

F = 280 kg * 9.8 m/s^2

Now, we can substitute the force and displacement values into Hooke's Law to find the spring constant:

k = F / x

Substituting the values:

k = (280 kg * 9.8 m/s^2) / 0.025 m

Calculating this expression gives us the stiffness of one spring.