Solution to problem 15.3.6 from the collection of Kepe O.E.

The problem considers a material point M with mass m, suspended on a thread of length OM = 0.4 m to a fixed point O. Initially, the point was set at an angle? = 90° from the equilibrium position and released without initial velocity. It is necessary to determine the speed of this point as it passes through the equilibrium position. The answer to the problem is 2.80.

The problem can be solved using the law of conservation of energy. When a point moves in a circle, its kinetic energy K and potential energy P are related as follows: K = P.

In the equilibrium position, the potential energy of the system will be maximum and equal to mgh, where h is the height of the point’s suspension, equal to 0.4 m, and g is the acceleration of gravity.

At the maximum deviation of a point from its equilibrium position, its potential energy will be zero. Consequently, its kinetic energy will be maximum and equal to mg(cos?), where cos? is the cosine of the angle of deviation of the point from the equilibrium position, and g is the acceleration of gravity.

Thus, the speed of the point at the moment of passing through the equilibrium position will be equal to the root of the expression 2gh, which is 2.80.

Solution to problem 15.3.6 from the collection of Kepe O.?.

This digital product is a solution to problem 15.3.6 from the collection of physics problems by Kepe O.?. The problem describes the motion of a material point M with mass m, suspended on a thread to a fixed point O and retracted to an angle? from the equilibrium position. The solution to the problem is carried out using the law of conservation of energy and allows us to determine the speed of the point at the moment of passing through the equilibrium position.

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Digital product - solution to problem 15.3.6 from the collection of problems in physics by Kepe O.?. The problem describes the motion of a material point M with mass m, which is suspended on a thread to a fixed point O and moved at an angle of 90 degrees from the equilibrium position without an initial speed. The goal of the problem is to determine the speed of a point as it passes through the equilibrium position, with the answer being 2.80.

The solution to the problem is carried out using the law of conservation of energy, which connects the kinetic and potential energies of a material point. In the equilibrium position, the potential energy of the system will be maximum and equal to mgh, where h is the height of the point’s suspension, equal to 0.4 m, and g is the acceleration of gravity. At the maximum deviation of a point from the equilibrium position, its potential energy will be zero, and the kinetic energy will be maximum and equal to mg(cos?), where cos? - cosine of the angle of deviation of the point from the equilibrium position.

Thus, the speed of the point at the moment of passing through the equilibrium position will be equal to the root of the expression 2gh, which is equal to 2.80. The solution to the problem is presented in a convenient and beautiful html format, which makes it easy to read and study the material. This digital product can be used as a study guide or to prepare for exams. It is intended for anyone interested in physics and problem solving.

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The product in this case is the solution to problem 15.3.6 from the collection of Kepe O.?.

The problem considers a material point of mass m suspended on a thread 0.4 m long to a fixed point O. Initially, the point is moved at an angle of 90° from the equilibrium position and released without an initial velocity. It is necessary to determine the speed of this point as it passes through the equilibrium position.

The answer to the problem is 2.80.

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