Answer to Question #137322 in Mechanical Engineering for Chinny

Question

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.

Answer 1

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

V= $\sqrt {2gH}$

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

g = 32.2 ft/s²

Therefore ,

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

Therefore Kinetic energy

$KE=\frac{1}{2} \times{m} \times{V^2}$

$KE= 94050903.13$ ft-lbf

KE= 120861.75 BTU