During the process from the release of the rod to the start of a uniform motion, Joule heat is gener

You're confusing work with energy. Some physical quantities can directly represent energy and the work done by a force: mgh, for example

As for"fs"It only represents the work done by F, and although it does involve the transfer and transformation of energy, if you take it directly as some kind of energy, you will lose.

That's where you got it wrong.

On the left you wrote q. So first of all, let's be clear, the idea of this problem you are dealing with is the conservation of energy. Then the physical quantity that appears in the whole equation must be some kind of energy with a clear price. The right way it has been given.

What can be used for FS is the kinetic energy theorem. The method you want to use should be: F-Mgssin30-ampere force to do the work = 1 2mv 2

Among them, the "work done by amperes" is the work of changing force, which requires the help of high numbers. Finally, i can be solved, which is also a variable, and q can be obtained further.

"mgh" can be understood both as the work done by gravity and as gravitational potential energy. The work done by pulling force is very limited.

The metal rod seems to have mass, so the kinetic energy and potential energy should be considered, because it is uniform, the voltage and current are stable, so the formula is directly set. The conservation of energy must be found carefully and completely, otherwise even the formula is not written correctly, and there is no one.

Your style is not right.

According to the conservation of energy, the gravitational potential energy of m is converted into m, the kinetic energy of m, the gravitational potential energy of m, and the internal energy, that is, the Joule heat produced on the rod, you can't go wrong by repeating the equation according to this relation.

1) Under the action of the external force with constant power, the metal rod first does variable acceleration motion, and then does uniform motion, at this time, the ampere force and f force are balanced, and the relationship between ampere force and velocity is deduced by Faraday's law, Ohm's law and ampere force formula, and then the velocity is solved by the equilibrium condition;

2) In time t, the work done by the external force f is pt, and the external force f and ampere force do work on the metal rod, and the work done by the metal rod to overcome the ampere force is obtained according to the kinetic energy theorem

For the existence of a force, there must be an obstacle.

Not necessarily. The work of ampere force is also partly used to reduce the increase in kinetic energy.

The conductor rod does the movement of cutting magnetic inductance lines, which is equivalent to a power supply to R, which is known by the conservation of energy, and the electrical work of the power supply is all used to heat the resistor! That is, the electrical work of the power supply is equal to the Joule heating on the resistor.

Secondly, the conductor rod is equivalent to the power supply, therefore, the non-electrostatic force of this power supply is the ampere force, so the work done to overcome the ampere force is equal to the electrical work of the power supply.

Remember: In the electromagnetic induction problem, the ampere force is the bridge between electrical energy and other forms of energy, and work = the conversion of energy.

For example: a power supply, with a guide rail, the guide rail is placed horizontally, a metal rod is placed on the guide rail, regardless of all friction, the current flowing through the conductor rod at this time is , then the ampere force must be b l, after the conductor rod moves for a certain distance, the electrical energy consumed in the circuit is divided into two parts, one part is electric heat, and one part is converted into the mechanical energy of the conductor rod (only kinetic energy at this time). At this point, the value of mechanical energy is equal to the work done by the ampere force.

When resistors work, they are dissipating electrical energy and dissipating heat. If the surface temperature exceeds 40 degrees, there is a feeling of heat. We all choose to work below their rated power.

Measure its working power, and if it does not exceed its rated power, it can work for a long time even in a hot state.

You use a resistor at 25V, and the current should not be greater than 1A according to its rated power. Then this 25W resistor is consuming 25W power, it will be hot, if the heat dissipation is poor, and the ambient temperature is high, it will also be hot and shorten the life. Therefore, the resistor with high power should pay attention to the heat dissipation space.

If the human body can feel the heat, it is estimated that the temperature is 60 degrees Celsius.

1 Total Joule heating refers to the heat emitted by the work done by the current, and its determining formula is the square of the current multiplied by the resistance.

The heat of the hot pan R1 and R2 is in parallel, and the heat of R1 and R2 is the square of the current on it multiplied by the resistance.

The Joule heating of the total circuit is the square of the total current multiplied by the total resistance. Take a pure resistive circuit as an example, since the pure resistance satisfies Ohm's law, u=u r:

R1 R2: Q1 = U 2 R1, Q2 = U 2 R2, Q Total = U 2(R1+R2) R1R2

The problem description is not clear, please rephrase your problem.