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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.

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]

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.

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.