(1) Let the speed of uniform upward movement of ab rod be V.

The electromotive force generated by the ab rod is E = BLV (L is the distance between the two guide rails, and the topic seems to be expressed by Z? )

According to the right-hand rule, the current direction in the ab rod (power supply) is from B to A.

At this time, the ab rod is subjected to gravity m 1*g (vertically downward), tensile force f, supporting force n (vertically upward along the guide rail), ampere force f (downward along the guide rail) and sliding friction force f (downward along the guide rail).

The cd rod is always fixed, which can be regarded as a resistance. It is connected in parallel with the lamp as an external circuit, and the ab rod is the power supply.

Because the lamp emits light normally, the current in the lamp is I lamp = P lamp quantity /U lamp quantity = 3/6 = 0.5a..

Then the current in the cd rod is IC = U lamp/R2 = 6/4 =1.5a.

So the current in the ab rod is IA = I lamp+IC = 0.5+1.5 = 2 a.

For the ab pole, there are

Ia=E / R 1=BLV / R 1

That is 2 = 2 * 1 * V/ 1.

So the speed of the ab rod is v =1m/s.

(2) According to the left-handed rule, the direction of current in cd can be judged. The direction of the ampere force on the cd rod is perpendicular to the cd rod obliquely upward (upward to the left), and the included angle between this ampere force direction and the horizontal direction is θ = 37 degrees, so the direction of the static friction force it receives is horizontal to the right.

The fc static = b * IC * l * cos θ = 2 *1.5 *1* cos 37 degrees = 2 *1.5 *1* 0.8 = 2.4 n.

(Note: The vertical component of the ampere force of the cd rod is smaller than the gravity of the rod and will not leave the track.)

(3) According to the meaning of the question, the constant force on the ab rod f = m1* g * sinθ+f+b * ia * l.

And f = f=μ * m 1*g cosθ.

f = m 1 * g *(sinθ+μ* cosθ)+b * ia * l = 1 * 10 *(0.6+0.5 * 0.8)+2 * 2 * 1 = 65438+。

From the relationship between energy conversion and conservation

F * x = m1* g * x * sinθ+(m1* v 2/2)+μ * m1* gcosθ * x+q total.

Where q is always the total calorific value of ab rod, cd rod and lamp when electrified.

Get14 * 4 =1*10 * 4 * 0.6+(1*12/2)+0.5 *1*10.

Q total = 15.5 joules

The resistance of the lamp is R lamp = U lamp quantity 2/P lamp quantity = 6 2/3 = 12 ohm.

When the cd rod and the lamp are connected in parallel to form an equivalent resistance, the resistance of this equivalent resistance is R = 12 * 4/( 12+4) = 3 ohms.

This equivalent resistance is equal to the current flowing through the ab rod, so the ratio of the heating power of the equivalent resistance to the heating power of the ab rod.

P / Pa=R / R 1=3 / 1

Then the ratio of the total calorific value of the equivalent resistance to the total calorific value of the ab rod.

Q/QA = p/PA = 3 (equal heating time)

That is QA = q/3.

Q+QA = the sum of Q.

Q+(q/3) = 15.5 joules.

Q = 1 1.625 Joule

And because the R lamp and Rc are connected in parallel, there is Q lamp /Qc=R2/R lamp = 4/ 12 = 1/3.

The total calorific value of cd rod QC = 3 q lamp is obtained

Through q lamp +QC = q

Q lamp +3 Q lamp = Q = 1 1.625 Joule.

Then the total calorific value of the lamp is Q lamp =11.625/4 = 2.90625 Joule ≈ 2.9 Joule.