Question

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?

Answer 1

Answer:

$F_1 = 4F$

Explanation:

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

Then,

$W = Fl$

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

$Fl = KE_f - KE_i$

$Fl = KE_f$

$Fl = \frac{1}{2}mv^2$

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

$F_1l=\frac{1}{2}m(2v)^2$

$F_1l = \frac{1}{2}m(4v^2)$

$F_1l=\frac{1}{2}m(4v^2)$

The relation between this two states is given by

$\frac{F_1l}{Fl} = \frac{\frac{1}{2}m4v^2}{\frac{1}{2}mv^2}$

$\frac{F_1}{F} = 4$

$F_1 = 4F$

Therefore the magnitude force is required to double the speed of te shell in the same gun is $F_1 = 4F$