Solving for Acceleration: Mass, Air Resistance, and Cliff Falls

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To find the acceleration of a 33 kg rock falling under the influence of gravity and air resistance, one must consider the forces acting on it. The gravitational force is calculated as the mass multiplied by the acceleration due to gravity, resulting in approximately 323 N. The net force acting on the rock is the difference between the gravitational force and the air resistance of 239 N, leading to a net force of 84 N. Applying Newton's second law, the acceleration can be determined by dividing the net force by the mass, yielding an acceleration of about 2.55 m/s². Understanding the forces and applying the correct formula is crucial for solving this problem accurately.
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A rock of mass 33 kg accidentally breaks loose from the edge of a cliff and falls straight down. The magnitude of the air resistance that opposes its downward motion is 239 N. What is the magnitude of the acceleration of the rock?

I tried a=239*33 and I got 7.24 but that was wrong. Please help. I don't know what else to do.
 
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What are the forces acting on the block?

What does Newton's 2nd law state?
 
Did you draw a free body diagram of the falling rock including all forces acting upon the rock?

Did you write out \sum \vec F=m\vec a? I'm sure if you do the above and sum the forces you'll get the correct answer.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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