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Cannon Shot based on velocity

  • Thread starter Joshua Lee
  • Start date
1. Homework Statement
You are in charge of a cannon that exerts a force 11200 N on a cannon ball while the ball is in the barrel of the cannon. The length of the cannon barrel is 2.18 m and the cannon is aimed at a 37.9 ◦ angle from the ground. The acceleration of gravity is 9.8 m/s 2 . If you want the ball to leave the cannon with speed v0 = 78.4 m/s, what mass cannon ball must you use? Answer in units of kg.

2. Homework Equations

3. The Attempt at a Solution
Used the kinetic energy equation to solve for mass
Then, used the second equation to find work.
I do not know how to get work to kinetic energy to plug into the mass solved equation.


Homework Helper
Gold Member
This looks like a problem that assumes that you are familiar with a principle known as the "work energy theorem". This theorem relates the total work done on a particle to the change in the particle's kinetic energy. Have you covered this principle in class? If you are using a textbook, check the index to see if it is covered in the book.
We have covered the theorem in class, but I don't see how that would apply. Would the change in energy be broken down into EF-E0 which would then be broken down into KE-K0?
I ended up solving the question using the v^2 equation in kinematics and F=ma. Does that tie into it all?


Homework Helper
Gold Member
You can work the problem using ##\sum \vec{F} = m \vec{a}##. If so, then you do not need to use any energy or work concepts. If you are going to use ##\sum \vec{F} = m \vec{a}##, then you should be sure that you think about all of the forces acting on the ball. (However, it could be that you would find that one of the forces is so much larger than the others that you can get an accurate enough answer by neglecting all but the largest force.)

From your list of "relevant equations" I assumed that you are trying to solve the problem using work and energy concepts. In this case, you only need the work energy theorem. You do not need to bring in ##\sum \vec{F} = m \vec{a}##. You don't need to break down the energy into different types of energy. The statement of the theorem only deals with work and kinetic energy.

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