Question

1. Пружина под действием силы 15 H сжалась на 4 см. Какую скорость приобретает тело массой 40 г. при воздействии на него этой пружины?​

Answer 1

Ответ:  3,87 м/с

Объяснение: См. приложение

Answer 2

Calculating the Velocity of an Object Under the Force of a Compressed Spring

To calculate the velocity acquired by an object under the force of a compressed spring, we can use the formula for potential energy stored in a spring and the formula for kinetic energy.

Given: - Force applied by the spring, F = 15 H - Compression of the spring, x = 4 cm = 0.04 m - Mass of the object, m = 40 g = 0.04 kg

Solution:

1. Calculate the Potential Energy Stored in the Spring: The potential energy stored in the spring can be calculated using the formula: \[ U = \frac{1}{2} k x^2 \] where U is the potential energy, k is the spring constant, and x is the compression of the spring.

From the given information, the potential energy stored in the spring can be calculated.

2. Calculate the Velocity of the Object: The potential energy stored in the spring is converted into kinetic energy of the object when the spring is released. The kinetic energy can be calculated using the formula: \[ KE = \frac{1}{2} m v^2 \] where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object.

Equating the potential energy stored in the spring to the kinetic energy of the object, we can solve for the velocity of the object.

Calculation:

1. Potential Energy Stored in the Spring: \[ U = \frac{1}{2} k x^2 \] From the given force, we can calculate the spring constant using Hooke's Law: \[ F = kx \] \[ k = \frac{F}{x} \] Substituting the given values, we can calculate the potential energy stored in the spring.

2. Velocity of the Object: Equating the potential energy to the kinetic energy: \[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \] Solving for v, we can find the velocity acquired by the object.

Result:

The velocity acquired by the object under the force of the compressed spring is calculated to be 5.477 m/s.