Physics

[woven wing]

I don't get how to do this problem

[marsh rapids]

Hi

15 meters for a barrel???

1250 kg for a Shell????

[sweet hill]

Hello

So first off force will be in Newtons (N) which is kg x m/s^2

so will want to keep units in all of your work and if you end up with Newtons at the end you can be fairly confident in your answer

[marsh rapids]

Okay but how can a bullet weight 1250 kg??

[sweet hill]

shell not bullet

[sweet hill]

The Mk. 8 APC (Armor-Piercing, Capped) shell weighed 2,700 lb (1225 kg) and was designed to penetrate the hardened steel armor carried by foreign battleships.[2][unreliable source?] At 20,000 yards (18 km) the Mk. 8 could penetrate 20 inches (508 mm) of steel armor plate.[20] At the same range, the Mk. 8 could penetrate 21 feet (6.4 m) of reinforced concrete.[20]

https://en.wikipedia.org/wiki/Armament_of_the_Iowa-class_battleship#Ammunition

[marsh rapids]

Dang!

So lets say the shell is a system of some explosive chemicals and a bullet.

The shell will explode causing the bullet to move at high speed outward the barrel

[marsh rapids]

You should first off write the fundamental principle of dynamics

Mass of the shell•acceleration of the shell= sum of all external forces acting on the shell.

There are two forces here:
the weight P and the explosive force of the explosion on the shell Fe

So m•a=P+Fe

[woven wing]

what if we just use change in
KE = W ==> kf - ki = F * d ==> (mv^2/2)/d = F(avg)

idk what we use as v tho, do we have to consider that it's 45degrees incline above the ground or no?

[marsh rapids]

Oh well I didn’t think about kinetic energy first hand

[marsh rapids]

v is the initial velocity of your shell right after the explosion

But with time the velocity will decrease

So if you plan on using kinetic energy then you should use ki=1/2mv_0^2

And yes you consider that it is 45 degrees inclined

Because when you use the work of a force, you use vectorial expressions

It has a direction

Not just a magnitude

So you have Ec=W(P)+W(Fe) with Fe the average force you’re looking for

[fickle wedge]

calculate the total amount of momentum in the shell and divite it by the time the shell was inside the barrel, but this would be if we didnt account for the gravity acting on the shell, but if thats the case you should just do some trig and you would find it