Question
What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2.40×104 m/s22.40 \times 10^4 \textrm{ m/s}^2? What is the magnitude of the force exerted on the ship by the artillery shell?
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Final Answer

2.6×107 N2.6 \times 10^7 \textrm{ N} is the force exerted on the shell due to the ship, which also is the magnitude of the force exerted on the ship due to the shell. Newton's Third Law reminds us that the magnitude of the force exerted by one object (ship, say) on another object (like a shell) is the same as the magnitude of the force exerted by the other object (shell) on the first object (ship).

Solution video

OpenStax College Physics, Chapter 4, Problem 15 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We're going to assume that the force exerted on the artillery shell by the boat is the net force on the artillery shell which isn't strictly true which is why I've drawn a squiggly sort of approximately equal sign here because there is also the earth exerting a force on the artillery shell. So this force exerted by the boat isn't the entire force, it's also the earth's force. But this force due to gravity is small in comparison to the force exerted by the boat because the force exerted by the boat is, you know, with this assumption we can calculate it to be 2.6 times ten to seven newtons. But the weight of the artillery shell being it's mass times gravitational field strength of the earth is 1100 kilograms and let's say times ten approximately and that's about eleven thousand newtons which is 1.1 times ten to the four newtons which is three orders of magnitude smaller than this force. So this weight would affect some decimal way over here somewhere after the six and -- but this number is precise only to two significant figures anyway, so this would have no effect that would make any difference on our calculation. So this assumption is okay. So there is a force 2.6 times ten to the seven newtons applied by the boat on to the artillery shell. The force applied on the boat is equal and opposite to the force applied on the artillery shell. So let's write that a bit better here. So the force on the boat due to the artillery shell is in the opposite direction to the force on the artillery shell due to the boat. These are Newton's third law pairs because they are the force that one body exerts on another. Well that body, that other thing will exert the same force on the original thing but in the opposite direction. All right. So, 2.6 times ten to the seven newtons in the opposite direction to the shell's acceleration.

Comments

Could you please share the explanation and answer to the second part of the question? Thank you.

Hi Alicia, I clarified in the answer that the force on the shell due to the ship is the Newton's Third Law counterpart to the force exerted on the ship due to the shell. These two forces have equal magnitudes, but they're in the opposite direction. The ship accelerates the shell in what we'll take to be the positive direction, whereas the shell accelerates the ship in the opposite direction with the same magnitude of force.
Hope this helps,
Shaun