Question

мальчик стреляет из пружинного пистолета вертикально вверх.Снаряд,имеющий массу 5 г,поднимается на высоту 40 м. Определите деформацию пружины этого пистолета до выстрела, если ее жесткость 200H/м.(4,47см)

Answer 1

Calculation of the Spring Deformation

To determine the deformation of the spring in the given scenario, we can use the principle of conservation of mechanical energy. The potential energy gained by the projectile as it rises to a height of 40 m is equal to the potential energy stored in the compressed spring.

The potential energy gained by the projectile can be calculated using the formula:

Potential Energy = mass * gravitational acceleration * height

Given: - Mass of the projectile (m) = 5 g = 0.005 kg - Height reached by the projectile (h) = 40 m - Gravitational acceleration (g) = 9.8 m/s^2

Using the formula, we can calculate the potential energy gained by the projectile:

Potential Energy = 0.005 kg * 9.8 m/s^2 * 40 m

Now, let's calculate the potential energy:

Potential Energy = 0.005 kg * 9.8 m/s^2 * 40 m = 1.96 J

Since the potential energy gained by the projectile is equal to the potential energy stored in the compressed spring, we can equate it to the formula for potential energy stored in a spring:

Potential Energy = (1/2) * spring constant * deformation^2

Given: - Spring constant (k) = 200 N/m - Deformation of the spring (x) = ?

Using the formula, we can calculate the deformation of the spring:

1.96 J = (1/2) * 200 N/m * x^2

Simplifying the equation:

0.98 J = 100 N/m * x^2

x^2 = 0.98 J / 100 N/m

x^2 = 0.0098 m^2

Taking the square root of both sides:

x = √(0.0098 m^2)

x ≈ 0.099 m

Therefore, the deformation of the spring in the pistol before the shot is approximately 0.099 meters.

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