Let the displacements be S 1, S2 and S3 respectively, and the corresponding times be t 1, t2 and t3 respectively.
S 1=0.9m v0=0？ uniformly accelerated motion
S2= 1m？ v 1=3m/s？ Uniform motion
S3= 1.6m initial velocity v 1=3m/s final velocity v3=5m/s uniformly accelerated motion.
( 1)S 1=0.9m v0=0？ uniformly accelerated motion
Through the formula v2=2as
Get: a 1=5m/s2,
t 1=v 1a 1=0.6s
t2=S2v 1= 13s
v32-v 12=2a3S3？
Solution: a3=5m/s2
t3=0.4s
T total =t 1+t2+t3=43s.
(2) When the wire frame passes through the magnetic field, the wire frame moves at a uniform speed and the force on the wire frame is balanced.
In AA'a'a area, analyze the stress of wireframe.
mgsinθ=ma 1
When passing through the magnetic field region,
F an =BIL=mgsinθ
BLBLv 1R=ma 1
According to stem analysis, the width of wireframe is L = D = S22 = 0.5m.
The solution is B=33T.
(3) Let the speed of entering the magnetic field above the metal frame under the constant force be V, and according to the kinetic energy theorem, we can get:
FS3-mgS3sinθ= 12mv2
When the wireframe passes through a magnetic field, F=mgsinθ+BLBLv? rare
Again? mgsinθ=ma 1
The solution is v= 163m/s and F=25 18N.
Answer: (1) The acceleration a of the metal frame before entering the magnetic field area is 5m/s2, and the time required to slide from the top to the bottom of the inclined plane is 43s.
(2) The magnetic induction intensity of uniform magnetic field is 33T.
(3) constant force f is 25 18N. ..