Determine the kinetic energy of a 4000-lbm body after it falls 1050 ft from rest.?
Determine the kinetic energy of a 4000-lbm body after it falls 1050 ft from rest. Assume that there are no heat and frictional effects, and that acceleration due to gravity is equal to 31.5 ft/s^2. *
- PhilomelLv 71 month ago
- NCSLv 71 month ago
@Jim is correct -- one DOES need to convert to slugs.
4000lbm / 32.2lbm/slug = 124 slugs
Then @Ash has it right (procedurally):
KE = mgh = 124slug * 31.5ft/s² * 1050ft = 4.11x10^6 lb·ft
The fundamental thing to remember when dealing with Imperial weights is that
1 lbm * g = 1 slug * 1 ft/s²
Then the rest is pretty easy.
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- oubaasLv 71 month ago
kinetic energy KE is equal to gravitational potential energy GPE , therefore :
KE = m*g*h = 4,000*31.5*1,050 ≅ 1.323*10^8 lb*ft
since 1 lb*ft is equal to 0.453 kg * 9.806 N/kg * 0.3048 m= 1.354 N*m or joule, then :
1.323*10^8 lb*ft = 1.323*10^8*1.354 = 1.79*10^8 joule
- AshLv 71 month ago
Conservation of energy
KE = PE
KE = mgh
KE = (4000 lbm)(31.5 ft/s²)(1050 ft)
KE = 1.323 x 10⁸ ft lb
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- JimLv 71 month ago
So you have PE = KE
mgh = ½mv²
but sorry, I cant work problems in the American system.
I think pounds have to be converted to slugs.