# 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. *

Please help

### 5 Answers

- 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.