Solution to problem 15.5.2 from the collection of Kepe O.E.

15.5.2 Four masses of mass m = 1 kg each, connected by a flexible thread thrown over a stationary weightless block, move according to the law s = 1.5t². Determine the kinetic energy of the load system at time t = 2 s. (Answer 72)

Given a system of four weights weighing 1 kg each, connected by a flexible thread, which is thrown over a fixed block. The motion of the system is described by the law s = 1.5t². It is necessary to find the kinetic energy of the load system at time t = 2 s.

To solve the problem, it is necessary to find the speed of the loads at time t = 2 s, using the derivative of the function s with respect to time. For ease of calculation, you can replace the formula s = 1.5t² to a more convenient formula v = 3t. Thus, the speed of the loads at time t = 2 s will be equal to v = 6 m/s.

Next, you can find the kinetic energy of each of the loads using the formula E_To = (mv²)/2 and then add the resulting values. Since all the loads are moving at the same speed, the kinetic energy of each of them will be the same and equal to E_To = (16²)/2 = 18 J. Thus, the kinetic energy of the load system at time t = 2 s will be equal to 418 = 72 J.

Solution to problem 15.5.2 from the collection of Kepe O.?.

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The solution to problem 15.5.2 describes the motion of a system of four weights thrown over a stationary block. The solution is presented in the form of a detailed description of the steps required to solve the problem, with explanations and comments. The solution also contains graphic and numerical illustrations that will help you understand the process of solving the problem and get the correct answer.

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Solution to problem 15.5.2 from the collection of Kepe O.?. is a digital product that is a detailed description of the solution to a physics problem. The problem considers the motion of a system of four weights weighing 1 kg each, connected by a flexible thread thrown over a stationary block. The motion of the system is described by the law s = 1.5t2. It is necessary to determine the kinetic energy of the load system at time t = 2 s.

To solve the problem, it is necessary to find the speed of the loads at time t = 2 s, using the derivative of the function s with respect to time. For convenience of calculations, you can replace the formula s = 1.5t2 with the more convenient formula v = 3t. Thus, the speed of the loads at time t = 2 s will be equal to v = 6 m/s.

Next, you can find the kinetic energy of each load using the formula Ek = (mv2)/2 and then add the resulting values. Since all the loads move at the same speed, the kinetic energy of each of them will be the same and equal to Ek = (162)/2 = 18 J. Thus, the kinetic energy of the load system at time t = 2 s will be equal to 4*18 = 72 J.

By purchasing the solution to problem 15.5.2 from the collection of Kepe O.?. in digital format, you receive a ready-made solution to the problem with explanations and comments, which can be useful for both students and teachers involved in physics.

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Solution to problem 15.5.2 from the collection of Kepe O.?. consists in determining the kinetic energy of a system of four loads at time t = 2 s, provided that the loads move according to the law s = 1.5t^2 and have a mass m = 1 kg each.

To solve the problem, you need to use the formula for the kinetic energy of a system of bodies:

Ek = (m1v1^2 + m2v2^2 + ... + mnvn^2) / 2,

where Ek is the kinetic energy of the system, m1, m2, ..., mn are the masses of bodies, v1, v2, ..., vn are the velocities of bodies.

First, it is necessary to determine the speed of the loads at time t = 2 s. To do this, we use the formula for speed:

v = ds / dt,

where ds is the movement of the body, dt is the time period.

Let us differentiate the given expression for the law of movement of loads to find the speed:

v = dv/dt (1.5t^2) = 3t.

Thus, the speed of each load at time t = 2 s will be equal to:

v = 3 * 2 = 6 м/c.

Let's substitute the obtained values ​​of mass and speed into the formula for kinetic energy:

Ek = (m1v1^2 + m2v2^2 + m3v3^2 + m4v4^2) / 2 = (1 * 6^2 + 1 * 6^2 + 1 * 6^2 + 1 * 6^2) / 2 = 72 hours.

Thus, the kinetic energy of the system of four weights at time t = 2 s is 72 J.

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