Physics 30 Jan 01 Diploma Exam

# Include calculations to support the values you write

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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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