Question #137322
Assuming that there are no heat effects and no frictional effects, find the kinetic energy (in Btu) and speed (in fps) of a 3,330-lb body after it falls 877 ft from the rest. Start with the steady flow equation, deleting energy terms which are irrelevant.
1
Expert's answer
2020-10-08T15:11:41-0400

The velocity of boby falling through height H when it started from rest is given as,

V= 2gH\sqrt {2gH}

Given as - H=877 ft, mass (m)= 3330 lb

g = 32.2 ft/s2

Therefore ,

V=2×32.2×877\sqrt {2 \times 32.2 \times 877} =237.67 m/s

Therefore Kinetic energy

KE=12×m×V2KE=\frac{1}{2} \times{m} \times{V^2}

KE=94050903.13KE= 94050903.13 ft-lbf

KE= 120861.75 BTU




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