No components are necessary, so on to Step 3 : write Newton’s second law for each block, calling
clockwise rotation of the pulley positive:
Step 4 : Solve algebraically. It’s easiest to add these equations together, because the tensions cancel:
mg = (3m )a , so a = g /3 = 3.3 m/s^2 .To get the tension, just plug back into T − mg = ma : T = m (a + g ) = (4/3)mg = 13 N.
Electric and Magnetic Fields
How to Do It
The force of an electric field is F = qE , and the direction of the force is in the direction of the field for a
positive charge. The force of a magnetic field is F = qvB sinθ , and the direction of the force is given by
the right-hand rule.
The Drill
The magnetic field above has magnitude 3.0 T. For each of the following particles placed in the field, find
(a) the force exerted by the magnetic field on the particle, and (b) the acceleration of the particle. Be sure
to give magnitude and direction in each case.
1 . an e- at rest
2 . an e- moving ↑ at 2 m/s
3 . an e- moving ← at 2 m/s
4 . a proton moving at 2 m/s
5 . an e- moving up and to the right, at an angle of 30° to the horizontal, at 2 m/s
6 . an e- moving up and to the left, at an angle of 30° to the horizontal, at 2 m/s
7 . a positron moving up and to the right, at an angle of 30° to the horizontal, at 2 m/s
8 . an e- moving → at 2 m/s
9 . a proton moving at 2 m/s