How Is Kinetic Energy Transferred from a Hammer to a Nail?

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Discussion Overview

The discussion revolves around estimating the kinetic energy transferred from a hammer to a nail during impact. Participants explore the calculations involved in determining kinetic energy, the assumptions necessary for such calculations, and the factors affecting energy transfer.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the kinetic energy of the hammer using the formula for kinetic energy, suggesting a value of 1237.5 J based on the hammer's mass and speed.
  • Another participant points out that while the hammer's kinetic energy can be calculated, the actual energy transferred to the nail cannot be determined without additional context, such as the conditions of the impact.
  • Concerns are raised about the use of grams instead of kilograms in the calculations, indicating a potential issue with unit consistency.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the amount of kinetic energy transferred to the nail, with some emphasizing the uncertainty due to various factors affecting the impact.

Contextual Notes

Limitations include the lack of information on the conditions of the hammer's impact with the nail, which affects the energy transfer, as well as the need for consistent units in calculations.

dg_5021
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Estimate the kinetic energy acquired by a 12g nail when it is struck by a 550g hammer moving with a speed of 4.5 m/s?

How do u find the kinetic energy? i first mutiply 550 x 4.5= 2475 /12= 206.25 am i doing it right?
 
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dg_5021 said:
Estimate the kinetic energy acquired by a 12g nail when it is struck by a 550g hammer moving with a speed of 4.5 m/s?

How do u find the kinetic energy? i first mutiply 550 x 4.5= 2475 /12= 206.25 am i doing it right?

Kinetic energy is

\frac {1}{2} m v^2

For your problem, you have to determine how much kinetic energy the hammer transfers to the nail.
 
1/2 550 x 4.5 = so then it will be 1237.5J right?
 
Work on those units chuck...
 
I think this is a very strange question. You can calculate the hammer's energy, but you can't know how much of it is transferred to the nail. Is somebody holding the hammer? Is somebody throwing it? Will it bounce?

What you've calculated is the hammer's energy, but if you want the answer in Joules you should use kilograms, not grams, as your unit for mass.
 

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