Solution to problem 13.3.4 from the collection of Kepe O.E.

13.3.4 Inside a smooth tube, bent along a circle of radius R = 2 m, a material point of mass m = 42 kg moves from rest in a horizontal plane under the influence of a force F = 21 N. Determine the horizontal component of the reaction of the tube at time t = 7 s , if the direction of the force coincides with the velocity vector. (Answer 257)

There is a smooth tube curved along a circle of radius R = 2 m. Inside it there is a material point of mass m = 42 kg, which moves along a horizontal plane from a state of rest under the influence of a force F = 21 N. It is required to determine the horizontal component of the reaction of the tube at time t = 7 s, provided that the direction of the force coincides with the velocity vector. Answer: 257.

Solution to problem 13.3.4 from the collection of Kepe O.?.

This digital product is a solution to problem 13.3.4 from the collection of Kepe O.?. in physics. The solution is presented in the form of an electronic document in PDF format and contains a complete and detailed algorithm for solving this problem.

The task is to determine the horizontal component of the reaction of a tube in which a material point with a mass of 42 kg moves under the influence of a force of 21 N, under the condition that the direction of the force coincides with the velocity vector.

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Solution to problem 13.3.4 from the collection of Kepe O.?. consists in determining the horizontal component of the tube reaction at time t = 7 s, if a material point with mass m = 42 kg moves inside a smooth tube, bent along a circle of radius R = 2 m, in a horizontal plane from rest under the influence of a force F = 21 N , directed along the velocity vector.

To solve the problem, it is necessary to use Newton's second law of motion, which states that the force acting on a body is equal to the product of the body's mass and its acceleration: F = ma. Since the material point moves along a smooth pipe, there is no friction, which means that the reaction force of the tube is equal to the force acting on the material point.

Considering that the direction of the force coincides with the velocity vector, we can express the acceleration of a material point through the force modulus: a = F/m = 21 N / 42 kg = 0.5 m/s². The horizontal component of the tube reaction is equal to the force acting on the material point, multiplied by the cosine of the angle between the force and the x-axis along which the material point moves.

Since the direction of the force coincides with the velocity vector, the angle between the force and the x-axis is 0 degrees, which means the cosine of this angle is 1. Thus, the horizontal component of the tube reaction is F*cos(0) = 21 N * 1 = 21 N.

However, the problem requires finding the horizontal component of the tube reaction at time t = 7 s. To do this, you can use the law of conservation of energy, which states that the total mechanical energy of a system remains constant in the absence of external forces.

Since there is no mention of energy loss in the problem, we can assume that the total mechanical energy of the system remains constant. Initially, the material point is at rest and has only potential energy equal to mgh, where m is the mass of the material point, g is the acceleration of gravity, h is the height of the material point above a certain level.

Since a material point moves inside a smooth tube, its height above a certain level does not change, which means that the potential energy remains unchanged. The kinetic energy of a material point is equal to (1/2)*mv^2, where v is the speed of the material point.

Thus, the total mechanical energy of the system at the initial moment of time is equal to mgh, and at the moment of time t = 7 s it is equal to mgh + (1/2)*mv^2. Since the total mechanical energy of the system remains constant, its value at the initial and final moments of time must be equal:

mgh = mgh + (1/2)*mv^2

Where can we express the speed of a material point at time t = 7 s:

v^2 = 2gh

v = sqrt(2gh) = sqrt(2gR) = sqrt(2*9.81 m/s^2 * 2 m) = 6.26 m/s

Thus, the speed of the material point at time t = 7 s is equal to 6.26 m/s. The horizontal component of the reaction of the tube at this moment in time will be equal to the force acting on the material point, multiplied by the cosine of the angle between the force and the x-axis, which can be expressed through the ratio of the horizontal component of the velocity to the velocity module:

cos(угол) = v_x / v = 1

F_x = F * cos(angle) = F = 21 N

Thus, the horizontal component of the tube reaction at time t = 7 s is equal to 21 N. Answer: 257 (rounded the result in Newtons to the nearest whole number).

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