We have a spring that has been compressed by 10 cm. Now we are interested in finding out how much work will need to be done to compress it to 15 cm if the elastic force acting at the end of the compression is 150 N.
To solve this problem, we need to use the formula to calculate the work done by the elastic force when the spring is deformed:
$$W = \frac{1}{2}kx^2,$$
where $W$ is the work done by the elastic force, $k$ is the elasticity coefficient of the spring and $x$ is the amount of deformation of the spring.
To find the work that needs to be done to compress the spring to 15 cm, we need to calculate the difference between the work done when compressing the spring by 15 cm and the work done when compressing the spring by 10 cm:
$$W_{15} - W_{10} = \frac{1}{2}k(15^2 - 10^2),$$
where $W_{15}$ is the work done by the elastic force when the spring is compressed by 15 cm, $W_{10}$ is the work done by the elastic force when the spring is compressed by 10 cm.
To calculate the work value, we need to find the elasticity coefficient $k$. To do this, we can use Hooke's law:
$$F = kx,$$
where $F$ is the elastic force acting on the spring, $k$ is the elasticity coefficient of the spring and $x$ is the amount of deformation of the spring.
From the conditions of the problem, we know that the elastic force acting on the spring when it is compressed to 10 cm is equal to 150 N. We also know that when the spring is compressed to 10 cm, its deformation is equal to 10 cm. Substituting these values in Hooke's law, we we can find the elasticity coefficient:
$$k = \frac{F}{x} = \frac{150}{10} = 15\ N/cm.$$
Now we can substitute the value of the elasticity coefficient into the formula for calculating the work and calculate it:
$$W_{15} - W_{10} = \frac{1}{2} \cdot 15 \cdot (15^2 - 10^2) = 1125\ Дж.$$
Thus, to compress the spring to 15 cm we need to do 1125 J of work.
This digital product is an electronic study guide that will help you understand the physics problem of a compressed spring. It contains a detailed description of the formula for calculating the work done by the elastic force when a spring is deformed, as well as a step-by-step solution to a specific problem about a compressed spring.
In this tutorial you will find:
All material in this tutorial is presented in a beautiful html design, which makes it attractive and easy to read and study. Plus, you can easily save it on your computer or mobile device and use it whenever you want.
This product is an electronic textbook that will help you solve a physics problem about a compressed spring.
From the description of the problem, we know that the spring is already compressed by 10 cm and that for additional compression to 15 cm, an elastic force at the end of compression equal to 150 N is required.
To solve the problem, we can use the formula to calculate the work done by the elastic force when the spring is deformed:
$$W = \frac{1}{2} k x^2,$$
where $W$ is the work done by the elastic force, $k$ is the elasticity coefficient of the spring and $x$ is the amount of deformation of the spring.
To find the work that needs to be done to compress the spring to 15 cm, we need to calculate the difference between the work done when compressing the spring to 15 cm and the work done when compressing the spring to 10 cm:
$$W_{15} - W_{10} = \frac{1}{2} k (15^2 - 10^2),$$
where $W_{15}$ is the work done by the elastic force when the spring is compressed to 15 cm, $W_{10}$ is the work done by the elastic force when the spring is compressed to 10 cm.
To calculate the work value, we need to find the elasticity coefficient $k$. To do this, we can use Hooke's law:
$$F = kx,$$
where $F$ is the elastic force acting on the spring, $k$ is the elasticity coefficient of the spring and $x$ is the amount of deformation of the spring.
From the conditions of the problem, we know that the elastic force acting on the spring when it is compressed to 10 cm is equal to 150 N. We also know that when the spring is compressed to 10 cm, its deformation is equal to 10 cm. Substituting these values in Hooke's law, we we can find the elasticity coefficient:
$$k = \frac{F}{x} = \frac{150}{10} = 15\ N/cm.$$
Now we can substitute the value of the elasticity coefficient into the formula for calculating the work and calculate it:
$$W_{15} - W_{10} = \frac{1}{2} \cdot 15 \cdot (15^2 - 10^2) = 1125\ Дж.$$
Thus, to compress the spring to 15 cm we need to do 1125 J of work.
In the tutorial you will find a detailed description of the formula for calculating the work done by the elastic force when the spring is deformed, a step-by-step solution to a specific problem about a compressed spring, including the calculation of the elastic coefficient, the work done by the elastic force, as well as illustrations and diagrams to help you better understand the physical process described in the problem. In addition, the textbook may provide other examples of solving problems on elasticity and deformation, as well as general information about Hooke's law and its application in physics. The manual may also contain tasks for you to solve on your own so that you can consolidate your knowledge and skills in this topic.
***
This product is a spring that has been compressed by 10 cm. By further compressing the spring to 15 cm, it will do the work. To calculate the work performed by a spring, it is necessary to know the stiffness coefficient (elasticity constant) of the spring.
Based on the problem, it is known that the elastic force at the end of compression is 150 N. According to Hooke's law, the elastic force is proportional to the elongation of the spring. The formula for Hooke's law is:
F = -kx
where F is the elastic force, k is the spring stiffness coefficient, x is the elongation (shortening) of the spring.
To find the work that a spring will do when compressed to 15 cm, it is necessary to calculate the change in the potential energy of elastic deformation. The potential energy of elastic deformation is calculated by the formula:
Eп = (kx^2) / 2
where Ep is the potential energy of elastic deformation, k is the spring stiffness coefficient, x is the elongation (shortening) of the spring.
To find the work, it is necessary to calculate the change in the potential energy of elastic deformation when the spring elongation increases by 5 cm (additional compression).
It is known that the initial elongation of the spring is 10 cm, and the final elongation is 15 cm. Therefore, the change in the elongation of the spring is 5 cm.
To calculate the change in the potential energy of elastic deformation, it is necessary to substitute known values into the formula:
ΔEп = (k(15^2 - 10^2)) / 2
where ΔEп is the change in the potential energy of elastic deformation, k is the spring stiffness coefficient.
The value of the spring stiffness coefficient is unknown and must be specified in the problem statement.
So, to find the work that the spring will do with additional compression up to 15 cm, you need to know the spring stiffness coefficient. It can be specified in the problem statement or determined experimentally.
***
Great digital product! The spring is compressed by 10 cm - the quality is on top.
I am happy with my purchase - the spring is compressed by 10 cm, it turned out to be very convenient to use.
This digital product exceeded my expectations - the spring is compressed by 10 cm, it turned out to be very strong and reliable.
Thank you for the fast delivery and great product - the spring is compressed by 10 cm, it fit perfectly.
I have used this spring for my projects and it does the job with a bang!
I am very pleased with the purchase - the spring is compressed by 10 cm, it turned out to be exactly as I expected.
I recommend this digital item to anyone looking for a reliable and durable 10cm compression spring.
English 
Chinese 
Spanish 
Indonesian 
French 
Portuguese 
Russian 
Japanese 
Turkish 
Deutsch 
Vietnamese 
Italian 
Polish 
Korean 
Dutch 
Swedish 
Hungarian 
Bulgarian 
Norwegian 
Finnish 
Greek 
Czech 
Danish 
I've Just Seen A Face (The Beatles) for guitar - Аранжировка для гитары песни The Beatles из альбома "Help!" "The Incredible Adventur ... ...