j/How much energy is re-
quired to raise the submerged
box through a height∆y?k/A seesaw.l/The biceps muscle is a
reversed lever.proportion to one another, so this has no effect on the conclu-
sion.)
Conservation of energy does not, however, tell us anything obvi-
ous about which ball gets there first. This is a general problem
with applying conservation laws: conservation laws don’t refer di-
rectly to time, since they are statements that something stays the
same at all moments in time. We expect on intuitive grounds that
the ball that goes by the lower ramp gets to B first, since it builds
up speed early on.
Buoyancy example 10
.A cubical box with massmand volumeV =b^3 is submerged
in a fluid of densityρ. How much energy is required to raise it
through a height∆y?
.As the box moves up, it invades a volumeV′=b^2 ∆ypreviously
occupied by some of the fluid, and fluid flows into an equal volume
that it has vacated on the bottom. Lowering this amount of fluid
by a heightbreduces the fluid’s gravitational energy byρV′gb=
ρgb^3 ∆y, so the net change in energy is
∆E=mg∆y−ρgb^3 ∆y
= (m−ρV)g∆y.In other words, it’s as if the mass of the box had been reduced by
an amount equal to the fluid that otherwise would have occupied
that volume. This is known as Archimedes’ principle, and it is
true even if the box is not a cube, although we’ll defer the more
general proof until page 207 in chapter 3. If the box is less dense
than the fluid, then it will float.
A simple machine example 11
.If the father and son on the seesaw in figure k start from rest,
what will happen?
.Note that although the father is twice as massive, he is at half
the distance from the fulcrum. If the seesaw was going to start
rotating, it would have to be losing gravitational energy in order to
gain some kinetic energy. However, there is no way for it to gain
or lose gravitational energy by rotating in either direction. The
change in gravitational energy would be
∆U=∆U 1 +∆U 2
=g(m 1 ∆y 1 +m 2 ∆y 2 ),but∆y 1 and∆y 2 have opposite signs and are in the proportion of
two to one, since the son moves along a circular arc that covers
the same angle as the father’s but has half the radius. Therefore
∆U= 0, and there is no way for the seesaw to trade gravitational
energy for kinetic.
Section 2.1 Energy 85