Answer to Question #256042 in Molecular Physics | Thermodynamics for Johanna

Question

Assuming that there are no heat effects and frictional effects. find the kinetic energy and speed of a 3220 lb body after it falls 778 ft from rest.

Answer 1

Assuming that there are no heat effects and no factional effects, find the kinetic energy and speed of a 3220-lb body after it falls 778 ft from rest. Start with the steady flow equation, deleting energy terms which are irrelevant.

$m=3220 \;lb \\ h=778 \;ft \\ u=0 \\ v^2=0^2+2gh \\ v^2 = 2gh \\ v = \sqrt{2gh} \\ g=9.8 \;m/s^2 = 9.8 \times 3.28 = 32.14 \;ft/s^2 \\ v = \sqrt{2 \times 32.14 \times 778} \\ v= 223.64 \;ft/s \\ v≈ 224 \; ft/s \\ Kinetic \; energy \;KE = \frac{1}{2}mv^2 \\ KE = \frac{3200 \times (224 )^2}{2 \times 32.2 } \\ KE = 2510827.376 \;lb \; ft \\ KE = \frac{2510827.376}{778} = 3227.285 \;BTU$