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DonReynolds

  • 2 years ago

A system of two block with one frictionless pulley is release. Block one mass 3.00kg sitting on ground, block 2 mass 5.00 kg with a height of 4.00 meter. What is the velocity (speed) of the 5.0kg block in m/s when it strikes the ground. We have been using the conservation of momentum and Kinetic Energy (KE) and Potential Energy (PEg) formulas. ow do I set this up to solve?

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  1. imron07
    • 2 years ago
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    Since there's no energy lost in the system: PE+KE before = PE + KE after |dw:1351478956312:dw|

  2. DonReynolds
    • 2 years ago
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    Thanks for that reply, I understand that, but how do I set up the equation to calculate the velocity?

  3. imron07
    • 2 years ago
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    Before: m1gh1+m2gh2+1/2 mv1^2+1/2 mv2^2=0+5g*4+0+0 After: m1gh'1+m2gh'2+1/2 mv'1^2+1/2 mv'2^2=3g*4+0+1/2x3xv'^2+1/2x5xv'^2 (v'1=v'=v')

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