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Unformatted text preview: A (J) 1.4 1.4 Mechanical Energy
of the magnet at
position B (J) 1.4 0.47 Resisting Force on
the magnet (N) 0 0.98 vi • Potential Energy
EpA = mgh
Ep A = (0.150 kg)(9.81 m/s2)(0.95 m)
= 1.398 J • Acceleration: from d = vi t +
aglass = 1
2 at 2 a= 2(0.95 m) aaluminum = (0.44s) 2 aglass = 9.814 m/s2 • Kinetic Energy
vf 2 = vi2 + 2ad
Ek = 1
2 mvf 2 2(0.95 m)
(0.76 s) 2 aaluminum = 3.289 m/s2 vf 2 = 2ad since vi = 0
Ek = 1
2 m(2a net d ) = manetd Eglass = manetd Eglass 2d
t2 Ealuminum = manetd = (0.150 kg)(9.8 m/s 2 )(0.95 m)
= (0.150 kg)(3.3 m/s 2 )(0.95 m)
= 1.397 J
Ealuminum = 0.470 J vi + vf
t . Since vi = 0, vf = 2d . As a result:
t
2
= 4.32 m/s, and vf ( a l u m i n u m ) = 2.50 m/s Students can calculate vf from d =
vf ( g l a s s ) vii • Resisting Force
Method 1
∆E = Work done = Fd
∆E
Fglass =
d
1.4 J − 1.4 J
=
0.95 m
Fglass = 0 N ∆E
d
1.4 J −0.47 J
=
0.95 m
= 0.979 N Faluminum = Faluminum Method 2 Fnet
Fg – FA
(0.150 kg)(9.81 m/s2) – FA
FA = ma
= ma
= (0.150 kg)(3.29 m/s2)
= 0.98 N • Lenz’s Law Explanation
By Lenz’s Law, a changing magnetic field will cause an induced magnetic field in
the aluminum tube that opposes the original changing magnetic field.
Since the glass tube is an electrical insulator, there is no induced magnetic field. viii...
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This note was uploaded on 01/14/2014 for the course PHYSICS Physics 30 taught by Professor Quinlan during the Fall '09 term at Centennial High School.
 Fall '09
 Quinlan
 Physics

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