Motion Question: Speed of 150kg Pile Driver Falling 7.5m

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Homework Help Overview

The problem involves a 150-kg pile driver falling from a height of 7.5 m and seeks to determine its speed upon impact with the piling. The context relates to concepts of motion and energy in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of energy conservation principles to analyze the problem. There is mention of specific formulas, including one derived from energy considerations, and questions about the applicability of these formulas in different contexts.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some guidance has been provided regarding energy conservation and relevant formulas, but there is no explicit consensus on the best method to apply.

Contextual Notes

There is a noted lack of coverage in the original poster's textbook regarding the relevant formulas, which may be impacting their understanding and ability to solve the problem effectively.

wickid
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Hello,

I have been racking my brains on this one, and I am sure it is an easy question for most. My teacher has been skipping chapters and jumping around in our text a lot and I cannot get the formula to solve this problem. If anyone can help me out with a formula I would be very appreciative. The question is as follows.

A 150-kg pile driver falls from a height of 7.5 m to hit a piling.

What is its spead as it hits the piling.

As I don't have the time it takes to hit the piling, I am unable to tell how fast the pile driver is falling.

I'm hoping this is an easy problem, and in reviewing the chapter I am in, I cannot find a sample problem similar to this, nor a formula that would work. Please help with with a good formula to use to solve this problem.
 
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Try viewing the problem in terms of the energy of the pile driver. The key equation, as in all conservation of energy problems, is

[tex]\Delta K = -\Delta U[/tex]


It is almost always easier to treat "constant acceleration problems" in terms of energy rather then fiddling and twisting the equations of motion in every direction.
 
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Thank you for your response,

My text doesn't seem to cover that forumula in the energy/work chapter. I did run into formula v=Sq Root(2gh) which would be about 12.1 m/s. When checking the answer in the book, it is 12 m/s but is this an accurate formula to use in the future, or a coinsidence in this problems case?
 
That equation will definitely work provided that gravity is the only force, you neglect air resistance. It is derived directly from energy considerations, like quasar recommended.
 
Try deriving it yourself, it follows from the premise that:

The net kinetic energy is a result of a change in potential energy. Algebraically:

[tex]\frac{1}{2} mv^2 = mgh[/tex]

Now solve for v! This works great for freefall (for speeds much less than the escape speed)

Also, solving for h, you can find the height that an object started freefalling from if you know its final velocity.

Energy is very useful when considering such events.
 
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