Solution to problem 20.6.3 from the collection of Kepe O.E.

In a given mechanical system, kinetic energy is expressed in terms of generalized velocities s1 and s2 as follows: T = 0.5 s12 + s22 + s1s2. The generalized forces in the system corresponding to these velocities are equal to QS1 = -3H and QS2 = 2H, respectively. It is necessary to determine the acceleration s2. The solution is to calculate the second derivative of kinetic energy with respect to s2, that is, a2 = d2T/ds22, where d2T/ds22 = 2. Replacing this value and the value of QS2 in the formula, we obtain: 2 = 2s2 - 3, whence s2 = 5. Thus, acceleration s2 is 5.

Solution to problem 20.6.3 from the collection of Kepe O.?.

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This digital product is a solution to problem 20.6.3 from the collection of problems in physics by Kepe O.?. The problem involves a mechanical system, kinetic energy, and generalized velocities. The product includes a detailed description of the problem, a step-by-step solution with detailed explanations and formulas, and a final answer.

To solve the problem, it is necessary to determine the acceleration s2, which requires calculating the second derivative of the kinetic energy with respect to s2, that is, a2 = d2T/ds22. The value of this derivative is known and equal to 2. Substituting the value of the generalized force QS2, which corresponds to the speed s2, we obtain the equation 2 = 2s2 - 3. Having solved it, we obtain the answer: acceleration s2 is equal to 5.

The solution is presented in a beautiful html format, which makes it convenient and enjoyable to read. By purchasing this digital product, you save your time and receive a ready-made solution to the problem, which can be used to prepare for exams, independently study a topic, or simply to expand your knowledge in the field of physics.

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Solution to problem 20.6.3 from the collection of Kepe O.?. consists in determining the acceleration s2 of a mechanical system, for which the kinetic energy T is expressed through the generalized velocities s1 and s2, and the generalized forces QS1 and QS2 are equal to -3H and 2H, respectively.

To solve the problem, it is necessary to use the equation of motion in generalized coordinates:

QS = d/dt(dT/ds') - dT/ds,

where QS is the generalized force, T is the kinetic energy, s is the generalized coordinate, t is time.

Let us differentiate the kinetic energy by the generalized speed s2:

d(T)/ds2 = s2 + s1

Further, according to the formula for the generalized force QS2:

QS2 = d/dt(dT/ds2) - dT/ds2

we get:

2 = d/dt(s2 + s1) - s2 - s1

Considering that s1 and s2 are functions of time, let us differentiate the equation again with respect to time:

0 = d2s2/dt2 - 1

Thus, the acceleration s2 is 1 m/s^2. Answer: 5 (m/s^2).

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