Mechanical Energy vs Potential Energy & Kinetic Energy

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Homework Help Overview

The discussion revolves around a problem involving mechanical energy, potential energy, and kinetic energy, specifically in the context of a hammer striking a block to ring a bell at a certain height. The scenario includes calculations related to energy transfer and the required speed of the hammer.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between kinetic energy and potential energy, questioning the calculations related to the hammer's required speed. There are attempts to clarify the energy transfer involved in the scenario.

Discussion Status

Some participants express confusion over the initial calculations, with one suggesting a revised speed that seems more reasonable. There is an ongoing exploration of the energy concepts involved, but no consensus has been reached.

Contextual Notes

Participants are working under the assumption of no energy losses in the system, which may influence their calculations and interpretations of the problem.

Lexington
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[SOLVED] Mechanical Energy vs Potential Energy & Kinetic Energy

I'm pretty sure I did this one wrong, =( please help clarify?

8. At the Calgary Stampede, you can win a prize if the bell rings when you strike a wooden block with a 10.0-kg hammer. The block is at one end of a lever. The other end of the lever drives a 2.0-kg metal bar up a slide to ring the bell 9.0 m above. What is the minimum speed the hammer must be swung to make the bar hit the bell?

Assuming there are no energy losses, the energy required to be transmitted to the 2.0-kg bar can be calculated and used as the kinetic energy required for the hammer.



Ek = 1/2mv^2
Ep = mgh
Em = Ep + Ek
v = √[(2•Ek)/m]




m1 = 10.0kg
m2 = 2.0kg
mt = m2 + m1 = 12.0kg
D = 9.0m

Ep = mgh
= (2.0kg)(9.0m)(9.81m/s^2)
= 176.58 J

Hammer:
h = 0m
v = √[(2•Ek)/m] + mgh
= √[(2•176.58J)/10.0kg] + (10.0kg)(9.81m/s^2)
= 104.042726646919
v = 1.0x10^2 m/s

The hammer should have a velocity of 1.0x10^2 J when it hits the wooden block, causing 1.77x10^2 J of energy on the 2.0kg block.
 
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Lexington said:
Hammer:
h = 0m
v = √[(2•Ek)/m] + mgh
= √[(2•176.58J)/10.0kg] + (10.0kg)(9.81m/s^2)*0
:wink:
 
Thank you! 5.9 m/s sounds allot better than 104m/s haha!
 
Lexington said:
Thank you! 5.9 m/s sounds allot better than 104m/s haha!
A pleasure :smile:
 

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