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Mechanical Energy vs Potential Energy & Kinetic Energy

  • Thread starter 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.
 

Answers and Replies

  • #2
Hootenanny
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Hammer:
h = 0m
v = √[(2•Ek)/m] + mgh
= √[(2•176.58J)/10.0kg] + (10.0kg)(9.81m/s^2)*0
:wink:
 
  • #3
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Thank you! 5.9 m/s sounds allot better than 104m/s haha!
 
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Thank you! 5.9 m/s sounds allot better than 104m/s haha!
A pleasure :smile:
 

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