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The force exerted by expanding gases is what propels a shell out of a gun barrel. Suppose a force that has an average magnitude F propels the shell so that it has a speed v at the instant it reaches the end of the barrel. Part A What magnitude force is required to double the speed of the shell in the same gun?

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Answer:

[tex]F_1 = 4F[/tex]

Explanation:

We need to find a proportion between the two force. This proportion is easly identificable through the Work made for both.

Then,

[tex]W = Fl[/tex]

The work done by the force on the shell = change in kinetic energy,

[tex]Fl = KE_f - KE_i[/tex]

[tex]Fl = KE_f[/tex]

[tex]Fl = \frac{1}{2}mv^2[/tex]

We know can calculate the speed of the gun when is doubled.

[tex]F_1l=\frac{1}{2}m(2v)^2[/tex]

[tex]F_1l = \frac{1}{2}m(4v^2)[/tex]

[tex]F_1l=\frac{1}{2}m(4v^2)[/tex]

The relation between this two states is given by

[tex]\frac{F_1l}{Fl} = \frac{\frac{1}{2}m4v^2}{\frac{1}{2}mv^2}[/tex]

[tex]\frac{F_1}{F} = 4[/tex]

[tex]F_1 = 4F[/tex]

Therefore the magnitude force is required to double the speed of te shell in the same gun is [tex]F_1 = 4F[/tex]

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