The guide rail makes uniform acceleration motion, and the initial velocity is zero, v=at,

E=BLat，

s= 12at2

Loop: ohm's law of closed circuit;

I = blvr total = blatr+2r0 (12at2) = blatr+r0at2.

(2) The guide rail is subjected to external force F, ampere force FA and friction force F, wherein

Amplitude force on the rod: FA=BIL=B2L2atR+R0at2.

ff =μFN =μ(mg+BIL)=μ(mg+B2 L2 ATR+r0at 2)

According to Newton's law, F-FA-Ff=Ma.

f = Ma+FA+Ff = Ma+μmg+( 1+μ)b2l 2 ATR+r0at 2

In the above formula, when: Rt=R0at

That is, when t=RaR0, the external force f takes the maximum value,

f max = Ma+μmg+ 12( 1+μ)b2l 2 arr 0，

(3) Let the moving distance of the guide rail in this process be s,

From the kinetic energy theorem, W =△Ek W = MAS.

Since the friction force Ff=μ(mg+FA),

So friction does work: w = μ GS+μ WA = μ GS+μ Q,

s=W？ μQμmg，

△Ek=Mas=W？ μQμmgMa

Answer: (1) The expressions of induced electromotive force and induced current with time in the loop are E=BLat, i = blatr+r0at2;

(2) After the time of aRR0, the pulling force F reaches the maximum value, and the maximum value of the pulling force F is Ma+μ g+12 (1+μ) B2L2 ARR0.

(3) What is the increase of kinetic energy of guide rail? μQμmgMa。