c*x
发帖数: 555 | 1 A block slides down from a frictionless hill of height h.
The hill starts horizontally and finishes horizontally.
Initial speed is v1 and final speed is v2, both pointing
horizontally to the right.
The potential energy-kinetic energy conversion gives:
m(v2)^2/2 - m(v1)^2/2 = mgh
However if one sits in a car moving at v0 in the opposite
direction horizontally, he will observe the velocities of
the block as:
v1' = v1 + v0
v2' = v2 + v0
Thus m(v2')^2/2 - m(v1')^2/2 = ... = mgh + m(v2 - v1)v0 != mgh | h***o
发帖数: 539 | 2 discussion problem in undergrade level?
the assumption behind m(v2)^2/2 - m(v1)^2/2 = mgh is that nothing
except gravity force does work to the block...
in the first frame of reference, the assumption holds because the
force given by hill, F is always perpendicular to block's velocity,
F dot V = 0. Only gravity force did work.
in the second frame of reference, F dot V no longer equals to 0...
F did work. that's why you get m(v2')^2/2 - m(v1')^2/2 != mgh.
The above qualitative explanation should
【在 c*x 的大作中提到】
: A block slides down from a frictionless hill of height h.
: The hill starts horizontally and finishes horizontally.
: Initial speed is v1 and final speed is v2, both pointing
: horizontally to the right.
: The potential energy-kinetic energy conversion gives:
: m(v2)^2/2 - m(v1)^2/2 = mgh
: However if one sits in a car moving at v0 in the opposite
: direction horizontally, he will observe the velocities of
: the block as:
: v1' = v1 + v0
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