Note that F 2 and F 3 are each larger than F 1 since FE =.
Therefore, F 2 + F 3 is also larger than F 1.
- C Torque is equal to rFsinθ. In this case, the angle is 90º, and sin(90º) is
equal to 1. The equation can therefore be simplified to rF, where F = mg. In
this case, then, we find that torque = rmg = (0.5 m)(0.2 kg)(10 N/kg) = 1
N·m.
- B Because the rocks are dropped from rest simultaneously and air resistance
is negligible, both rocks will accelerate at the same rate and have the same
speed, v, at impact. So, the ratio of Rock 1’s momentum to Rock 2’s
momentum will be (M 1 v)/(M 2 v) = M 1 /M 2.
- B Because the forces N and w are vertical while the displacement of the
block is horizontal, the work done by each of these forces is zero, so they
contribute nothing to the total work performed. This eliminates (C), (D), and
(E). Since the force f is opposite to the direction of the displacement, the
work it does is negative; in fact, it’s −fL. The work done by the force F is
FL, so the total work performed on the block is −fL + FL = (F − f)L.
- A Choices (B), (C), (D), and (E) all change the magnetic flux through the
loop containing the light bulb, thus inducing an emf and a current. However,
just swinging the handle over as described in (A) changes neither the area
presented to the magnetic field lines from the bottom coil nor the density of
