Solution to problem 13.7.6 from the collection of Kepe O.E.

Consider elevator car 2 moving upward with acceleration a2 = 0.5g. It is necessary to determine the tension force of the spring if load 1, which weighs 100 N, is at rest relative to the cabin.

To solve this problem, it is necessary to apply Newton's second law, which states that the sum of all forces acting on a body is equal to the product of the body's mass and its acceleration. Thus, we can write the equation:

ΣF = m1 * a2 + Fн = m1 * 0

where ΣF is the sum of all forces acting on load 1, m1 is the mass of load 1, a2 is the acceleration of elevator car 2, Fn is the spring tension force.

Since the load is at rest relative to the cabin, its acceleration is zero. You can also take into account that the acceleration of gravity g is equal to 9.8 m/s².

Solving the equation, we get:

Fn = m1 * a2 = 100 N * 0.5 * 9.8 m/c² = 490 N

Thus, the spring tension is 490 N or 150 kg.

Solution to problem 13.7.6 from the collection of Kepe O..

This digital product is a solution to problem 13.7.6 from the collection of problems in physics edited by Kepe O. The problem is an example of calculating the tension force of a spring when a load is suspended on it and is at rest relative to a moving elevator car.

The solution to the problem was carried out by a professional physics teacher and checked for the correctness of the calculations. The material is presented in PDF format and can be downloaded to any device with Internet access.

By purchasing this digital product, you receive a ready-made solution to the problem, which can be used to independently prepare for exams in physics, as well as as additional material for improving your skills and expanding your knowledge in the field of physics.

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This product is a solution to problem 13.7.6 from the collection of problems in physics, edited by Kepe O.?. The problem is to determine the tension force of a spring when a load is suspended on it, which is at rest relative to a moving elevator car moving upward with an acceleration a2 = 0.5g. The solution to the problem was carried out by a professional physics teacher and checked for the correctness of the calculations.

This product is presented in PDF format and can be downloaded to any device with Internet access. By purchasing this product, you receive a ready-made solution to the problem, which can be used for independent preparation for exams in physics, as well as as additional material for expanding knowledge in the field of physics.

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Solution to problem 13.7.6 from the collection of Kepe O.?. consists in determining the tension force of the spring if a suspended load 1 weighing 100 N is in a state of relative rest, and the elevator car 2 moves upward with an acceleration a2 = 0.5g.

To solve the problem it is necessary to use Newton's laws and Hooke's law. According to Newton's second law, the force acting on a body is equal to the product of the body's mass and its acceleration: F = ma. According to Hooke's law, the elastic force acting on a spring is proportional to its elongation: F = kx, where k is the elasticity coefficient of the spring, x is its elongation.

In this problem, the load is in a state of relative rest, which means that it is acted upon by a force of gravity equal to its weight: F1 = m1g = 100 N, where m1 is the mass of load 1, g is the acceleration of gravity.

The tension force of the spring is directed upward and is equal to the sum of the forces acting on it: Fpr = F2 + F1, where F2 is the force with which the elevator car 2 acts on the spring.

According to Newton's first law, the sum of forces acting on a body is equal to the product of its mass and acceleration: ΣF = ma. In our case, the body is the system “load 1 + spring + elevator car 2”. Considering that the load is in a state of relative rest, and the cabin moves with acceleration a2, we obtain: Fpr - F1 = m2a2, where m2 is the total mass of the spring and the cabin.

Expressing the tension force of the spring from the last equation and substituting the value of gravity, we obtain: Fpr = m2a2 + F1 = m2a2 + m1g = (m2 + m1) a2. Thus, the tension force of the spring is equal to the sum of gravity and the force with which the cabin acts on the “load 1 + spring” system: Fpr = (m2 + m1) a2 = (m2 + m1) 0.5g = 150 N.

Answer: The tension force of the spring is 150 N.

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