Question

| 4.Исследуя свойства игрушечного пистолета, произвели выстрел пулей массой 30 г в горизонтальном направлении. Скорость пули в момент вылета 2 м/с. На сколько была сжата пружина в момент выстрела, если ее жесткость равна 0,3 кН/м? [4] ​

Answer 1

Ответ: 2см

Объяснение:

Дано:

m=30г=0,03кг

V=2м/с

k=0,3кН/м=300Н/м

----------------

x-?

Согласно закона сохранения энергии:

kx²/2=mV²/2

kx²=mV²

x=√(mV²/k)=√(0,03*2²/300)=0,02м=2см

Answer 2

Calculation of the Compression of the Spring

To calculate the compression of the spring in the moment of the shot, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the compressed spring is converted into the kinetic energy of the bullet as it is propelled forward.

The formula for the potential energy stored in a spring is given by:

Potential Energy = (1/2) * k * x^2

Where: - k is the spring constant (also known as the stiffness or the force constant) in N/m. - x is the compression or extension of the spring in meters.

In this case, we are given the mass of the bullet (30 g) and the velocity of the bullet at the moment of the shot (2 m/s). We are also given the spring constant (0.3 kN/m).

To find the compression of the spring, we need to rearrange the formula for potential energy and solve for x:

Potential Energy = (1/2) * k * x^2

Since the bullet is shot horizontally, we can assume that there is no change in gravitational potential energy. Therefore, the initial potential energy of the spring is equal to the kinetic energy of the bullet:

Potential Energy = Kinetic Energy

(1/2) * k * x^2 = (1/2) * m * v^2

Substituting the given values:

(1/2) * (0.3 kN/m) * x^2 = (1/2) * (0.03 kg) * (2 m/s)^2

Simplifying the equation:

0.15 * x^2 = 0.06

x^2 = 0.06 / 0.15

x^2 = 0.4

Taking the square root of both sides:

x = √0.4

x ≈ 0.6325 m

Therefore, the spring was compressed by approximately 0.6325 meters at the moment of the shot.