Answer:
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Explanation:
Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:
[tex]F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})[/tex] (1)
Where:
[tex]F[/tex] - Explosive force, measured in newtons.
[tex]\Delta s[/tex] - Barrel length, measured in meters.
[tex]m[/tex] - Mass of the shell, measured in kilograms.
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the shell, measured in meters per second.
If we know that [tex]m = 1250\,kg[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 750\,\frac{m}{s}[/tex] and [tex]\Delta s = 15\,m[/tex], then the explosive force experienced by the shell inside the barrel is:
[tex]F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}[/tex]
[tex]F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}[/tex]
[tex]F = 23437500\,N[/tex]
The explosive force experienced by the shell inside the barrel is 23437500 newtons.