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Re: Work-energy worries



"Lemmerhirt, Fred" wrote:

Would some object to this sort of question?

<facetious>
Yes.

It's a corollary of Murphy's law. You could assert that
2 and 2 makes 4 and somebody would take exception :-).
</facetious>

A 75kg boater tosses a 5kg anchor horizontally, straight forward across the
bow of his 125kg boat at 2m/s relative to the water. a) Neglecting any
horizontal force applied to the boat by the water, calculate the boat's
backward speed across the water just as the anchor leaves the boater's hand.
(The boat is initially at rest on the water.) b) At least how much work
must the boater have done to throw the anchor in this way?

Seriously, anybody should be able to figure out what was
intended by this question and come up with the canonical
answer.

There are multiple definitions of work, but only one that
IMHO makes sense, and (unsurprisingly) it makes sense in
this case. The other definitions, such as the one that
is connected to the execrable W+Q equation, don't make
sense in general and certainly don't make sense in this
case. Furthermore, I doubt anybody with the least bit
of sense would even attempt to use the W+Q formula in
this example, so the potential for confusion is minimal.

Nitpickers might feast on the following nits: The canonical
answer assumes, reasonably enough, that the anchor can be
treated as a "point particle". On the other side of that
coin, if we imagine that the anchor was initially rapidly
spinning, it would be theoretically possible to eject it
from the boat without doing any work, strictly speaking.

Also the canonical answer assumes that to a sufficient
approximation we can neglect the mass of the boater's
arm and hand relative to the other masses in the problem.
Nitpickers are invited to calculate the size of the
correction term.

This posting is the position of the writer, not that of Cambridge,
Masham, or Northumberland.

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.