The relationship between work and energy and displacement, force and displacement

You just have to remember that the work done by friction and heat is relative displacement.

As for this, it is possible to be on the ground or on other objects.

For example, if a wooden board slides on the ground, then the work done by friction is the displacement of the ground, which can also be said to be the relative displacement, but this relative object is very special to the ground.

If the plank slides on the plank and the plank below is also sliding, then the work done by friction between the two planks is the relative displacement between the two planks, and the work done by the frictional force between the plank below and the ground is the opposing displacement of the plank.

Let's use the simplest model, two planks are stacked on top of each other M1 and M2, the friction between the two is F1, the friction between the lower M2 and the ground is F2, and the upper plank M1 is affected by the tensile force of F, after a period of time, the position of M1 is shifted to S1, and the position of M2 is S2, and the relative displacement of the two is S=S1-S2

Let's start with a breakdown.

m1 is affected by the opposite f (to the right) and f1 (to the left), according to the kinetic energy theorem.

f*s1 - f1*s1 = ek1 - 0

M2 is affected by the opposite F1 (to the right, because F1 is the force between two planks, according to the Ox Three Theorem, both are reversed, the same magnitude) and F2 (to the left) is affected according to the kinetic energy theorem.

f1*s2 - f2*s2 = ek2 -0

Superimpose the above two formulas and you have it.

f*s1 - f2*s2 - f1*(s1-s2) = (ek1 + ek2) -0

This is the kinetic energy theorem for a system of two plates.

fs1 is the work done by the external kinetic force, f2*s2 is the work dissipated by the external resistance, and f1 (s1-s2) is the work done by the internal friction.

Therefore, in the case of a single analysis, the displacement to the ground is used, and the internal relative motion displacement is considered as a whole.

The condition of work is to experience a force and move some distance in the direction of this force.

If the object above and the object below are relatively stationary, then the frictional force does not do work and there is no energy converted into heat energy.

The displacement of an object to the ground refers to the displacement of the object with respect to the ground and can also be referred to as absolute displacement. Whereas, the displacement of the force when it does work on an object is not necessarily the displacement of the ground, it can be the displacement relative to other objects or the frame of reference.

For example, when a person uses force to push an object, the displacement of the object can be either relative to the ground or relative to other objects. For example, if an object is placed on a pulley system, the displacement of the object may be relative to the pulley, not to the ground.

Therefore, the displacement of the force when it does work on the object is not necessarily the displacement to the ground, it can be the displacement relative to other objects or the frame of reference.

The displacement of an object to the ground refers to the displacement of an object in the ground frame of reference and is usually used to describe the motion of an object on the ground of an oak ruler. For example, when an object is subjected to a force in the horizontal direction, the object moves along the horizontal direction and has little to do with the ground.

In physics, the direction of displacement of the work done by the force on the object depends on the direction of the force, if the force coincides with the direction of motion of the object, then the direction of the work done is the displacement to the ground, and vice versa. Of course, in many cases, the direction of the force on the object tends to be the same as the direction of the displacement towards the ground, such as swimming, so it can also be said that the displacement is the displacement of the ground when the force does work on the object.

The displacement of an object to the ground refers to the distance that an object moves relative to the ground, and this displacement is usually based on the horizontal direction and can be used to measure the movement of an object relative to the ground. However, when a force is applied to an object, the displacement of the object is not necessarily the displacement of the ground, as the pin may move in the vertical direction, or in other directions. Thus, when the force does work on an object, the displacement may or may not be a displacement to the ground.

The only exception to this is when an object moves in the same direction as the horizontal, its displacement is the displacement to the ground. Overall, understanding and differentiating between different types of displacement is very important for both mechanics learning and practice.

The displacement of an object to the ground refers to the displacement of an object with respect to the ground.

When the force does work on the object, the object will be displaced. However, the displacement is not necessarily a displacement to the ground. For example, when a person lifts an object in the air, the object is displaced, but there is no displacement to the ground.

Also, if all the forces experienced by an object are perpendicular to the ground, then the displacement of the force experienced by it to the ground is zero. For example, if a box is placed flat on the ground, if you push it hard, although the box is displaced, because the support force is perpendicular to the ground, the displacement of the force on the object is also zero.

In short, when calculating the displacement of the force on the object, it should be said that the direction of motion and the force on the object should be taken into account in order to calculate correctly.

d Analysis: According to the image, it can be seen that the external force on the object is changing, so it cannot be a uniform motion or a uniform velocity motion, and the answer can be known by the elimination method as dAccording to the simple harmonic movement.

It can be seen that the force is proportional to the displacement and the direction is opposite, so the image represents the force of the simple harmonic motion.

First of all, your teacher is right about the two ways to be sure. However, it is too close to the specific problem, and it is a bit of a model of the problem of dismantling the simple car, and there are many things that need to be memorized.

Actually, don't understand the two situations you mentioned that way, just remember the basic concepts: force, and the force has displacement, it is meritorious.

In question 1, for the frictional force of a, this force has a displacement, (note when understanding: it is the displacement of the force, not the displacement of the object), then it is a big deal.

In problem 2, the force of the baffle is acting on the spring, and this force has no displacement, so there is no work. But the spring has a displacement of the elastic force of the object above, and it is active.

Both force and displacement are vector quantities. Work is the internal product of force and displacement.

is a scalar quantity. 1) where is the angle between the force vector and the displacement vector.

For this equation to be correct, the force must be a constant vector and the path must be a straight line.

If the force changes with time or the path is not linear, the above equation no longer applies, in which case the line integral is used. Therefore, the general formula of merit is:

2) where c is the path; is a force vector; is a displacement vector. The expression is an inappropriate differentiation, path-related, and cannot be obtained by finding the differentiation.

Non-zero force can do no work, which is related to impulse.

Different. Impulse is the accumulation of force over time. The impulse is a vector quantity, so the centripetal force is present when moving in a circle.

No work is done, but a non-zero impulse is produced on the object.

The work done by the moment can be calculated from the following equation:

where is the moment.

Why the work done by friction on a slider and a block of wood is that the frictional force is multiplied by their displacement relative to the ground and not by the relative displacement of two objects.

Isn't the friction force between these two objects, and not between the object and the ground, but it should be the sliding friction multiplied by the displacement it produces.

When it comes to doing work, it is necessary to make clear which force is doing work on which object, what work is done (positive or negative), and how much work is done.

In the process, the analysis of the force and displacement shows that the slider is affected by the sliding friction of the wooden board to the left, and the slider (to the ground) displacements to the right, and the displacement magnitude is s+l, so the friction force does negative work on the slider, w1=-f*(s+l).

For the plank, it is subjected to the sliding friction of the slider f'To the right (f'and f is a pair of action reactions), the position of the board is shifted to s, so the friction force does positive work on the board, w2=f's=fs；

The algebraic sum of the work done by this pair of frictional forces is w=w1+w2=-fl

From the above analysis, it can be seen that when it comes to doing work, it is necessary to indicate which force does the work on which object. As for your doubts, it is just a misunderstanding: "it should be the sliding friction force multiplied by the displacement it produces", there is no "should" in physics, only the understanding and application of physical laws.

Energy is converted by doing work, such as when an object falls, work is done by gravity, and gravitational potential energy is converted into kinetic energy.

Force is everywhere, as long as it is a problem about force, it is necessary to use force analysis, and displacement can be well applied as long as a certain frame of reference is selected (the displacement record is a position change, so it is necessary to select the positive direction before it can be used, in fact, as long as the movement can be used displacement, just to see if it will simplify the problem and choose not to use it).

The work done by force measures the amount of change or transfer of energy, to be precise, because energy is conserved.

Force and displacement are the elements of work, and work is done when displacement occurs under the action of force, and there is a specific definition in the book.

The situation you said is that the work done is zero, and the work is a vector, which is equivalent to if you lift and put down a bucket of water, and the bucket is still in place, in fact, you have done the work, but the state of the bucket has not changed, and the work done twice before and after is zero.

You are mistaken. In the first question, the f-force is a variable force, and if you look at a circle from a microscopic perspective, the tangent direction of each of its points is it.

So the f-force is always consistent with the direction of motion of the circle, if you have to.

w=fs to explain it from a microscopic point of view, it is the superposition of countless small displacements.

So it does the work.

In the second question, the f force is a constant force, and the constant force can be expressed as w=fs, where s is the displacement if you are not good at that.

Comprehension can be understood in terms of energy relations, and the reduction in energy of a small object can only be done by the support force.

Negative work (it transfers energy to a part of the inclined plane).

W=flcosq f is a constant force (the magnitude and direction must be constant) l is the displacement.

1. First of all, I will explain to you what displacement is.

Work is equal to the force multiplied by the displacement, the force does not need to be explained, the displacement refers to the distance traveled in the direction of the force.

For example, in Question 1, the force is along the tangent direction of the application point, and the distance traveled is the distance along the tangent direction of the application point.

The distance is relative to the direction of the force.

In Question 2, the force exerted on the small block is perpendicular to the inclined plane, so the distance refers to the displacement along the inclined plane.

2 in your doubts2

The formula w=flcos , l refers to the length of the inclined plane, when the small object slides down, the inclined surface has a force along the inclined plane upward, but the small object slides downward, opposite to the upward component of the inclined plane support force. Therefore, the oblique does negative work on small blocks.

The displacement of a small block moving under the action of the component force in the horizontal direction of the inclined support force is 0 (ground, the inclined plane is smooth). So the work done by the force in the horizontal direction is 0

The total work is the work done by the force in the vertical direction of the inclined plane supporting the force.