Can you find the Force without acceleration?

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The discussion revolves around calculating the distance a cannonball rolled down a hill under the influence of gravity, given the work done on it. The key equation used is W = Fd cos(30), where W is the work done, F is the force, and d is the distance. Participants emphasize the importance of understanding forces acting on the cannonball, including gravitational and normal forces, and suggest using conservation of energy principles to simplify the problem. The conservation of mechanical energy is highlighted as a straightforward approach to relate the work done to the vertical distance traveled. Ultimately, the conversation underscores the significance of applying basic physics concepts and trigonometry to solve the problem effectively.
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Homework Statement


In 1453, during the siege of Constantinople, the Turks used a cannon capable of launching a stone cannonball with a mass of 5.40 X 10^2. Suppose a soldier dropped a cannonball with this mass while trying to load it into the cannon. The cannonball rolled down a hill that made an angle of 30 degrees with the horizontal. If 5.20 X 10^4 J of work was done by gravity on the cannonball on the cannonball as it rolled down a hill, how far did it roll?


Homework Equations


W=Fdcos(30)


The Attempt at a Solution


d=W/Fcos(30)
F=ma
d=W/macos(30)
d=(5300)/(540)acos(30)

What am I doing wrong?
 
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Draw a free body diagram. What forces are acting on the cannonball as it rolls down the hill?
 
I can't really draw on the screen but I can say it. The forces acting I guess are Fg(force of gravity)(down), the Fn (Normal force)(up) and the force of the ball Rolling downhill?(east)
 
You're making this much too difficult. Imagine a triangle making an angle 30 degrees off the horizontal. What does conservation of energy tell you about how much vertical distance is covered if a certain amount of energy is given to the object?
 
I don't know. Al we learned so far is the Conservation of Mechanical Energy, which is MEi =MEf, but to use this seems extremely complicated, unless I am misunderstanding it...
 
xZerocopyx said:
I don't know. Al we learned so far is the Conservation of Mechanical Energy, which is MEi =MEf, but to use this seems extremely complicated, unless I am misunderstanding it...

Conservation of energy is the simplest thing in the world! The energy a system is always conserved. The amount of energy gained by moving an object in a gravitational field in this situation is simply \delta PE = mg\delta hwhere m is the mass, g is the acceleration of gravity, and \delta h is the distance the object is moved. So if an object has had a certain energy put into it by a gravitational field (and you assume the object doesn't have any kinetic energy before and after), then you can tell how far it fell. Then it comes down to simple trigonometry.
 
If the only force acting on the object is the force of gravity, you can use g as the acceleration
 
Ok. thanks
 
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