what is the energy of the spring?
___
potential energy of the spring?
2. A body with a mass of 5 kg is located at a height of 12 m above the ground. Calculate its potential energy:
a) relative to the surface of the earth;
b) relative to the roof of the building, the height of which is 4 m.
___
3. An undeformed dynamometer spring was stretched by 10 cm, and its potential energy became 0.4 J. What is the coefficient of spring stiffness?
100 N, and to the second, with stiffness k2, - 50 N. How do the spring stiffnesses compare?
1) convert to si 2.5 t 350mg 10.5g 0.25t 2) it is necessary to determine the spring stiffness of the dynamometer if the distance betweendivisions 0 and 1 of its scale is 2 cm.
k=......................
what is the value of the force of gravity acting on the load
G=............................
3) for this task, you need a complete solution to determine the weight of an astronaut with a mass of 100 kg, first on the moon and then on Mars
4) it is necessary to determine the absolute elongation of the spring with a stiffness of 50 N/m if
it is acted upon with a force of 1 n and b) a body of mass 20 g is suspended from it
5) an astronaut, being on the moon, hung a wooden bar with a mass of 1 kg from a spring. the spring lengthened by two cm. then the astronaut, using the same spring, evenly pulled the bar along the horizontal surface. in this case, the spring lengthened by 1 cm
to be determined
spring stiffness ........................
the magnitude of the friction force ..........
how many times the friction force could be greater if the experiment was carried out on Mars
plz need in 4 hours i beg you
6. What is the stiffness of the spring if a force of 2 N stretched it by 4 cm?7. If the length of the spiral spring is reduced by 3.5 cm, an elastic force equal to 1.4 kN occurs. What will be the elastic force of the spring if its length is reduced by 2.1 cm?
8. When opening the door, the length of the door spring increased by 0.12m; the elastic force of the spring is at the same time 4 N. For which elongation of the spring is the elastic force equal to 10 N?
9. A force of 30 N stretches the spring by 5 cm. What force will stretch the spring by 8 cm?
10. As a result of stretching an undeformed spring 88 mm long, up to 120 mm, an elastic force equal to 120 N arose. Determine the length of this spring when the force acting on it is 90 N.
he is in balance.
Instruction
Pay attention to the fact that when a force arises, seeking to restore the original and shape of the given body. This force is caused by electromagnetic action that occurs between the atoms and molecules of the substance from which the spring is made. This force is called the elastic force. The simplest form - and compression.
Related videos
note
Be very careful when handling the spring. She, trying to straighten up, at any moment can "shoot" in an unpredictable direction and cause injury.
Strength gravity- this is such a force that shows the degree of attraction of the body to the Earth under the force of its gravity. It directly depends on body weight. Calculate strength gravity easy enough.
Related videos
note
1) The value of g varies slightly in different parts of the world. For example, in Moscow g = 9.8154 m/s², and in Cairo g = 9.79317 m/s².
2) It is worth noting that the acceleration of free fall (as a consequence, and the force of gravity) depends on:
- Masses of the planet;
- The radius of the given planet;
- The height of the body above the surface of the planet;
- Geographic location on the planet - on Earth, at the equator g = 9.78 m/s², and at the pole 9.83 m/s²;
- From the presence of minerals. For example - iron ore, which has pronounced magnetic properties.
Useful advice
1) Despite the fact that g is a constant value, g = 9.81 m/s² is used for technical calculations.
2) If we are talking about measuring the weight of a body, then numerically it is equal to the force of gravity.
3) Often, when the body is located on some kind of support, the so-called "resistance force of the support" is taken into account. It is directly proportional to the force of gravity, and also depends on the physical characteristics of the support itself, to which the force of gravity of a given body is applied.
Springs- this is a component of the car's suspension, which protect the car not only from road roughness, but also provide the desired body height above the road, which greatly affects the vehicle's handling, comfort and load capacity. As a result of tests for each car, the optimal rigidity suspension springs for certain driving conditions.
Instruction
When "breakdowns" occur, the suspension is considered too soft. In such situations, motorists become unstable in. Ideally, the spring force should be equal to a value that prevents excessive body roll.
Stiffer springs are required by cars that are prepared for racing. In different types of races of the same car, it involves installing springs with different rigidity Yu. When passing any turns, pay attention to body roll, which, with properly selected springs, should be no more than two or three degrees.
For the front and rear suspension, select springs according to stiffness in pairs. However, it is not immediately possible to achieve the desired suspension height, because the spring shrinks and can “lose” at the moment, which is very bad. This is due to a lack of bearing capacity even at full compression, but with rigidity yu, providing the desired height of the suspension. It is always easy to determine: between the coils of the spring there should be a gap of less than 4 mm.
Choose springs so that when filled, the gap between the coils of the springs is slightly more than 6.5 mm. It is advisable to install the softest springs, although they will roll the car within acceptable limits. It is usually incorrect to use stiff springs, relying on the fact that they reduce the roll of the car, improving handling.
Check rigidity springs according to the product code or according to the marks applied (by stamping or paint). Also define rigidity springs can be using hand, floor scales and a measuring ruler in kilograms per centimeter.
A wooden block (thickness not less than 12 mm) of a larger end area of the spring is placed on household floor scales, and a spring is installed on top. Then a second piece of wood and the length of the spring are placed on top of the spring. Using a press, the spring is compressed to a certain value (for example, 30 mm) and the readings of the scales are taken, thereby calculating rigidity.
note
The pressing force on the spring is measured according to the readings of the scales, but this method of determining the stiffness of the springs is dangerous, since the spring can fly off a fairly large distance.
What motorists go to to improve the quality of the ride. Among the many ingenious tricks, there is also a change in the ground clearance inherent in the design of the car. This can be done by changing the size of the screw springs shock absorber, that is, simply speaking, cutting it. You can carry out such a "surgical intervention" on your own. The main thing is to think carefully about the consequences of such an operation.
You will need
- - angle grinder ("Bulgarian");
- - hacksaw for metal;
- - a set of car wrenches.
Instruction
Deciding to perform pruning by force, first free springs by removing the stand. Support each side of the car in turn with a jack. Disconnect the wheels. Remove the bolts that secure the bottom of the rack. After that disconnect springs. Fold all fasteners carefully in one place, after cleaning them of dirt.
Decide how much you need to cut springs. For this, consult a car service specialist. To significantly change the clearance, you will need to cut one and a half to two turns. When in doubt, shorten first springs one turn and try them out. If necessary, the procedure can be repeated. cutting off springs for more turns at once, you, of course, will not be able to restore them to the required level later, so think carefully before taking up the tool.
Direct cutting of metal springs produce with the help of an angle grinder ("grinder"). If it is not available, use a hacksaw. Pre-markup in the right place. Cutting should be marked at the top of the product. This will reduce the negative consequences of the deformation of the updated springs.
Repeat the same operations for all springs, trying to end up with all of them being the same size. It is especially important that the size of the undercut springs coincides with the axes of the car in order to prevent loss of controllability due to even minimal distortion of the structure.
To avoid gross errors, use the capabilities of an automotive service department to trim the springs. A qualified one will allow you to assess how desirable it is for your vehicle, and will perform it at the highest professional level. Inept trimming of the springs may require their complete replacement in the future, and, consequently, unforeseen financial costs.
Related videos
Repairing furniture at home is a difficult task, but doable. On the one hand, a furniture specialist is guaranteed to fix a broken sofa. On the other hand, the delivery of overall furniture to the repair shop is a very troublesome and expensive task. With a strong desire, you can repair the sofa at home on your own.
You will need
- - wrench;
- - flat screwdriver;
- - pliers;
- - a knife for foam rubber;
- - new foam.
Instruction
Before proceeding with the repair, prepare the necessary space. To repair as quickly and efficiently as possible, prepare all the necessary tools, as well as a set of new replacement springs. Remove the carpet or carpeting, put a thick and dense film to protect the parquet from accidental dropping of a heavy tool or sofa structural element.
Start dismantling the sofa by detaching the sides. Unscrew the fasteners with a wrench, holding them as far as possible from free fall.
In any convenient way, mark the details of the folding mechanisms of the sofa planes on the right and left in order to assemble them in the correct order after repair. Dismantle mechanisms.
Continue disassembling the sofa, alternately disconnecting the seat, backrest. If there is a pallet, you need to dismantle it too.
To remove the covers, use pliers and pull out the fastening brackets. Avoid strong jerks, try to do everything carefully. Otherwise, you can damage or even tear the fabric of the covers.
Proceed to repair the spring block, after pulling out the mattress. Take a damp rag and remove the dust and wood shavings that are usually present in abundance at the place where the spring block is attached to the bottom of the sofa.
Test all sofas springs. This can be done without dismantling, however, experts still recommend pulling out the entire block and determining the suitability for further use of each spring separately.
Remove broken springs with pliers. Having installed new ones, use furniture foam rubber and an old thick blanket for greater elasticity of the spring block and increase the service life. To do this, take a knife, cut the foam rubber and stuff the spring block with it, then cover it with a blanket and fix it around the perimeter with a hammer and 30-40 mm nails so that the blanket “sits” as tightly as possible and collects the springs, preventing them from falling into different sides. Thus, the main cause of breakage or excessive stretching of sofa springs is eliminated - their non-synchronous operation.
Sources:
- how to change springs on mercedes
Energy is a physical concept that accompanies any movement or activity. This parameter in a conventionally closed system is a constant value regardless of the interactions between bodies occurring in it.
Instruction
Any movement or direct interaction of physical bodies is accompanied by the release, absorption or transfer of mechanical energy. The elements (bodies) of a mechanical system can either be in motion or at rest. In the first case, they talk about kinetic energy, in the second - about potential. Together, these quantities make up the total mechanical
We have repeatedly used a dynamometer - a device for measuring forces. Now let's get acquainted with the law that allows you to measure forces with a dynamometer and determines the uniformity of its scale.
It is known that under the action of forces arises body deformation– changing their shape and/or size. For example, an object can be molded from plasticine or clay, the shape and dimensions of which will be preserved even after we remove our hands. Such deformation is called plastic. However, if our hands deform the spring, then when we remove them, two options are possible: the spring will completely restore its shape and dimensions, or the spring will retain residual deformation.
If the body restores the shape and/or dimensions that it had before the deformation, then elastic deformation. The resulting force in the body is elastic force subject to Hooke's law:
Since the lengthening of a body is included in Hooke's law modulo, this law will be valid not only for tension, but also for compression of bodies.
Experiences show: if the elongation of the body is small compared to its length, then the deformation is always elastic; if the elongation of the body is large compared to its length, then the deformation will, as a rule, be plastic or even destructive. However, some bodies, such as rubber bands and springs, deform elastically even with significant changes in their length. The figure shows more than two times the extension of the dynamometer spring.
To clarify the physical meaning of the stiffness coefficient, we express it from the formula of the law. We obtain the ratio of the modulus of elasticity to the modulus of elongation of the body. Recall that any ratio shows how many units of the numerator are per unit of the denominator. That's why the stiffness coefficient shows the force that arises in an elastically deformed body when its length changes by 1 m.
- The dynamometer is...
- Due to Hooke's law, the dynamometer observes...
- The phenomenon of deformation of bodies is called ...
- We call a body plastically deformed, ...
- Depending on the modulus and/or direction of force applied to the spring, ...
- The deformation is called elastic and is considered to be subject to Hooke's law, ...
- Hooke's law is scalar in nature, since it can only be used to determine ...
- Hooke's law is valid not only for tension, but also for compression of bodies, ...
- Observations and experiments on the deformation of various bodies show that ...
- Ever since children's games, we know well that...
- Compared to the zero stroke of the scale, that is, the undeformed initial state, on the right...
- To understand the physical meaning of the stiffness coefficient, ...
- As a result of expressing the value "k" we...
- We know from elementary school mathematics that...
- The physical meaning of the stiffness coefficient is that it ...
In physics for grade 9 (I.K. Kikoin, A.K. Kikoin, 1999),
a task №2
to chapter " LABORATORY WORKS».
The purpose of the work: to find the stiffness of the spring from measurements of the elongation of the spring at different values of gravity
balancing force of elasticity based on Hooke's law:
In each of the experiments, the stiffness is determined at different values of the elastic force and elongation, i.e., the conditions of the experiment change. Therefore, to find the average stiffness value, it is not possible to calculate the arithmetic mean of the measurement results. We will use a graphical method for finding the average value, which can be applied in such cases. Based on the results of several experiments, we plot the dependence of the modulus of elasticity F control on the modulus of elongation |x|. When constructing a graph based on the results of the experiment, the experimental points may not be on a straight line that corresponds to the formula
This is due to measurement errors. In this case, the graph must be drawn so that approximately the same number of points is on opposite sides of the straight line. After constructing the graph, take a point on the straight line (in the middle part of the graph), determine from it the values of the elastic force and elongation corresponding to this point, and calculate the stiffness k. It will be the desired average value of the spring stiffness k cf.
The measurement result is usually written as an expression k = = k cp ±Δk, where Δk is the largest absolute measurement error. From the algebra course (VII class) it is known that the relative error (ε k) is equal to the ratio of the absolute error Δk to the value of k:
whence Δk - ε k k. There is a rule for calculating the relative error: if the value determined in the experiment is the result of multiplying and dividing the approximate values included in the calculation formula, then the relative errors add up. In that work
Means of measurement: 1) a set of weights, the mass of each is equal to m 0 = 0.100 kg, and the error Δm 0 = 0.002 kg; 2) a ruler with millimeter divisions.
Materials: 1) tripod with clutches and foot; 2) coil spring.
Work order
1. Attach the end of the coil spring to the tripod (the other end of the spring is equipped with an arrow pointer and a hook - fig. 176).
2. Next to or behind the spring, install and secure a ruler with millimeter divisions.
3. Mark and write down the division of the ruler against which the spring pointer falls.
4. Hang a weight of known mass from the spring and measure the extension of the spring caused by it.
5. To the first load, add the second, third, etc. weights, each time recording the lengthening |x| springs. According to the measurement results, fill in the table:
6. Based on the measurement results, build a graph of the dependence of the elastic force on the elongation and, using it, determine the average value of the spring constant k cp.
7. Calculate the largest relative error with which the value of kav was found (from the experiment with one load). In formula (1)
since the error in measuring the elongation Δx=1 mm, then
8. Find
and write your answer as:
1 Take g≈10 m/s 2 .
Hooke's law: "The elastic force that occurs when a body is deformed is proportional to its elongation and is directed opposite to the direction of movement of body particles during deformation."
Hooke's law
Rigidity is the coefficient of proportionality between the elastic force and the change in the length of the spring under the action of the force applied to it. According to Newton's third law, the modulus of the force applied to the spring is equal to the elastic force that has arisen in it. Thus, the stiffness of the spring can be expressed as:
where F is the force applied to the spring, and x is the change in the length of the spring under its action. Measuring instruments: a set of weights, the mass of each is equal to m 0 = (0.1 ± 0.002) kg.
Ruler with millimeter divisions (Δх = ±0.5 mm). The procedure for performing the work is described in the textbook and does not require comments.
|
weight, kg |  |
elongation |x|, | ||
Instruction
Attach a dynamometer to the body and pull it, deforming the body. The force that the dynamometer will show will be equal in absolute value to the elastic force acting on the body. Find the stiffness coefficient using Hooke, which says that the elastic force is directly proportional to its elongation and is directed in the opposite direction to deformation. Calculate the stiffness coefficient by dividing the value of the force F by the elongation of the body x, which is measured with a ruler or tape measure k=F/x. To find the elongation of a deformed body, subtract the length of the deformed body from its original length. Stiffness coefficient in N/m.
If there is no dynamometer, hang a load of known mass from the deformable body. Make sure that the body is deformed elastically and does not collapse. In this case, the weight of the load will be equal to the elastic force acting on the body, the stiffness coefficient of which must be found, for example, . Calculate the stiffness coefficient by dividing the product of the mass m and the gravitational acceleration g≈9.81 m/s² by the elongation of the body x, k=m g/x. Measure the elongation according to the method proposed in the previous one.
Example. Under a load of 3 kg, a spring 20 cm long became 26 cm, determine it. First find the extension of the spring at . To do this, from the length of the elongated spring, subtract its length in the normal state x=26-20=6 cm=0.06 m. Calculate the stiffness using the appropriate formula k=m g/x=3 9.81/0.06≈500 N / m.
And now a few tips. To reduce rigidity water in your , add distilled or clean rainwater to it, use special plants, such as elodea and hornwort. In addition, water can be frozen or boiled well. In the first case, it is poured into a low basin and exposed to frost. As soon as it freezes to half the capacity, the ice is broken through and, having melted, is used. In the second, water is boiled in enameled water for an hour, after which it is allowed to cool and two thirds of the “top” are used. water.
Related videos
As a result of the deformation of the physical body, a force always arises that opposes it, trying to return the body to its original position. Define this strength elasticity in the simplest case, according to Hooke's law.
Instruction
Strength elasticity, acting on a deformed body, arises as a result of the electromagnetic interaction between its atoms. There are different types of deformation: /stretching, shear, bending. Under the influence of external forces, different parts of the body move differently, hence the distortion and force elasticity, which is directed towards the previous state.
Tensile/compressive deformation by the direction of an external force along the axis of the object. It can be a rod, a spring, and another body that has a long shape. When distorted, the cross section changes, and the force elasticity is proportional to the mutual displacement of body particles: Fcontrol = -k ∆x.
This is called Hooke's law, but it is not always applied, but only for relatively small values of ∆x. The value k is called stiffness and is expressed in N/m. This coefficient depends on the initial material of the body, as well as the shape and dimensions, it is proportional to the cross section.
During shear deformation, the volume of the body does not change, but its layers change their relative to each other. Strength elasticity is equal to the product of the coefficient elasticity in shear, which is directly dependent on the cross section of the body, by the angle between the axis and the tangent, in the direction of which the external force acts: Fupr = D α.
