
John K. answered 03/16/16
Tutor
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Math and Engineering Tutor, Professional Engineer
As above kinetic and potential (static) are different measures of energy. It is useful to transform them. Suppose one launches a mass m (lbf-sec^2/ft) at 10 ft/sec from level ground. How high will the mass go above the ground.
The initial kinetic energy will converted to potential energy at the maximum height (v=0) as the energy is conserved. We have
1/2mv^2=mgh. To check the units are (lbf-sec^2/ft)*(ft^2/sec^2) or lbf-ft for kinetic energy and also(lbf-sec^2)*(ft/sec^2)*ft or ft-lbf for potential energy.
As 1/2mv^2=mgh we have v^2/2=gh or h = v^2/(2*g). If we approximate the earth's gravitational constant as 32.16 Then for our problem we have h= 10^2/(2*32.16)= 1.6 (ft^2/sec^2)/(ft/sec^2)
= 1.6 ft.