How Is the Average Force on a Shell Calculated Using the Work Energy Theorem?

AI Thread Summary
The discussion focuses on calculating the average force acting on a 25.8-kg shell fired at a speed of 880 m/s using the work-energy theorem. The kinetic energy acquired by the shell is expressed as (1/2)mv², leading to the equation Fd = (1/2)mv². A participant clarifies that the distance 'd' is 6.00 m, which is the length of the muzzle. This results in an average force of 1.66 MN. The conversation emphasizes understanding the application of the work-energy theorem in this context.
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Homework Statement



A weapon fired a 25.8-kg shell with a muzzle speed of 880 m/s. What averageforce acted on the shell?

Homework Equations



work energy theorem?

The Attempt at a Solution



Can someone please explain where the 6.00 comes from?

Starting from rest, the shell acquired a kinetic energy of (1/2)mv2

by a force, F,acting through a distance, d.Fd = (1/2)mv2
F(6.00) = (1/2)(25.8)(880)2
F = 1664960 N = 1.66 MN
 
Last edited:
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welcome to pf!

hi razzmatazz! welcome to pf! :wink:

6 m must be the length of the muzzle (didn't the question say so?) :smile:
 
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