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Solution
MECHANICS. 29I
train during this 40 seconds was 25 X 9750 = 243,750 foot-
pounds per second, and the horse-power exerted by the locomo-
tive on the draw-bar in starting this train was -VW— — 443.18
horse-power, but the power required to keep this train in motion
at a speed of 50 feet per second on a level road will be only
50 x 100°
= 90.91 horse-power. From this it may be seen what an
immense power there has to be produced in order to start heavy
machinery in a short time, in comparison to the power required
to keep it going after it is started.
Example 2.
How far did the train move before it got up to the required
speed of 50 feet per second ?
- 50X40 = 1000 feet.
2 2
Example 3.
Suppose that after the train had acquired this speed of 50
feet a second, the locomotive was detached and that the resist-
ance due to friction continued to be 1000 pounds. How many
seconds would the train be kept in motion by its momentum on
a level road ?
Solution
:
225400
Time = v m — X ~32~X
R — 1000 = 350 seconds.
Example 4.
How many foot-pounds of kinetic energy is stored in this
train, which weighs 225,400 pounds and runs at a constant speed
of 50 feet a second ?
K = v X m = o0 X 32.2 — 8,750,000 foot-pounds.
2 2
Example 5.
How far will this kinetic energy drive the train on a hori-
zontal road if we suppose the constant resistance due to friction,
as in Example 3, to be 1000 pounds?
Solution
:
Distance = ^tic_energy = 8750000
= sm feet
resistance 1000
When a body free to move is acted upon by a constant
force the space passed over increases as the square of time.
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