Solution to problem 15.2.9 from the collection of Kepe O.E.

In this problem, we consider load 2, which performs free vibrations in accordance with the law x = 0.1 sin 10t. The stiffness of spring 1 is 100 N/m. It is necessary to calculate the potential energy of the load at x = 0.05 m if its potential energy is zero at x = 0.

To solve this problem, we use the formula for the potential energy of the spring system:

U = (k * x^2) / 2,

where k is the spring stiffness, x is the displacement from the equilibrium position.

Based on the problem conditions, x = 0.05 m, and k = 100 N/m. Substituting the values ​​into the formula, we get:

U = (100 * 0.05^2) / 2 = 0.125 J.

Thus, the potential energy of the load at x = 0.05 m is 0.125 J.

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This product is a solution to problem 15.2.9 from the collection of Kepe O.?. in physics. The problem considers load 2, which oscillates freely according to the law x = 0.1 sin 10t. The stiffness of spring 1 is 100 N/m. It is necessary to determine the potential energy of the load at x = 0.05 m, if at x = 0 its potential energy is zero.

The solution to the problem is carried out using the formula for the potential energy of the spring system: U = (k * x^2) / 2, where k is the spring stiffness, x is the displacement from the equilibrium position. Based on the problem conditions, x = 0.05 m, and k = 100 N/m. Substituting the values ​​into the formula, we get: U = (100 * 0.05^2) / 2 = 0.125 J.

Thus, the potential energy of the load at x = 0.05 m is equal to 0.125 J. The presented digital product includes a complete and detailed solution to the problem, performed by qualified specialists in the field of physics. It will help you quickly and efficiently solve the problem, as well as significantly save time and effort on searching for information on your own. The product is designed in a convenient HTML format, which makes it easy to find the necessary data and formulas. By purchasing this product, you will improve your knowledge in the field of physics and be able to successfully cope with this task.

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I present a description of the solution to problem 15.2.9 from the collection of O. Kepe:

Hopefully:

• Load 2 oscillates freely according to the law x = 0.1 sin 10t.
• The stiffness of spring 1 is 100 N/m.
• x = 0 when the potential energy of the load is zero.
• It is required to determine the potential energy of the load at x = 0.05 m.

Answer:

1. Let's find the maximum value of the displacement of the load from the equilibrium position: x_max = 0.1 m.

2. Let's find the period of oscillation: T = 2π/ω, where ω = √(k/m), k is the spring stiffness, m is the mass of the load. m = 2 g, because cargo 2. ω = √(100/2) = 10 rad/s. T = 2π/10 = π/5 s.

3. Let's find the speed of the load at x = 0.05 m: v = dx/dt = 0.1*cos(10t)*10 = 1 m/s (since at x = 0.1 m the speed is zero).

4. Let's find the position of the load at time t: x = 0.1*sin(10t).

5. Let's find the potential energy of the load: Ep = kx^2/2, where k is the spring stiffness. At x = 0, the potential energy of the load is zero, so the change in potential energy is: ΔEп = Ep - 0 = k(x^2 - 0)/2 = 100*(0.05^2)/2 = 0.125 J.

Answer: the potential energy of the load at x = 0.05 m is 0.125 J.

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