anonymous
  • anonymous
A 4700 kg block falls vertically through 6.9 m and then collides with a 380 kg pile, driving it 3.3 cm into bedrock. Assuming that the block-pile collision is completely inelastic, find the magnitude of the average force on the pile from the bedrock during the 3.3 cm descent.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Michele_Laino
  • Michele_Laino
the kinetic energy of the falling block has to be equal to its potential energy, so we can write: \[\frac{1}{2}m_B{v^2} = m_Bgh\] where \(m_B=4700\), \(h=6.9\). So we can compute the magnitude of the velocity of the block when it collides with the pile: \[v = \sqrt {2gh} \] Now, since such collision has to be inelastic, then the work done by the resistance force \(F\), has to be equal to the kinetic energy of the system block+pile, so we can write: \[\frac{1}{2}\left( {{m_B} + {m_P}} \right){v^2} = Fd\] where \(m_P=380\), and \(v=\sqrt{2gh}\), \(d=0.033\,meters\). So we get: \[F = \frac{{\left( {{m_B} + {m_P}} \right){v^2}}}{{2d}} = \frac{{\left( {{m_B} + {m_P}} \right)}}{{2d}}2gh = \frac{{\left( {{m_B} + {m_P}} \right)gh}}{d}\]
Michele_Laino
  • Michele_Laino
please substitute numeric data into my formula above
Michele_Laino
  • Michele_Laino
please I have made an error, the quantity \(v\) is not \(\sqrt{2gh}\). Since the collision is inelastic, total momentum is conserved, so we can write: \[{m_B}{v_0} = \left( {{m_B} + {m_P}} \right){v_1}\] where I called \(v_0=\sqrt {2gh}\). So the speed of the obkect "block+pile" is: \[{v_1} = \frac{{{m_B}{v_0}}}{{{m_B} + {m_P}}}\] then the right formula for \(F\), is: \[F = \frac{{\left( {{m_B} + {m_P}} \right)v_1^2}}{{2d}} = \frac{{\left( {{m_B} + {m_P}} \right)}}{{2d}} \cdot {\left( {\frac{{{m_B}{v_0}}}{{{m_B} + {m_P}}}} \right)^2} = ...?\]

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anonymous
  • anonymous
Thank you , sorry I haven't responded earlier @Michele_Laino
anonymous
  • anonymous
is it 8960156.983 N?
Michele_Laino
  • Michele_Laino
yes! correct! I got: \(F=8,910,318.54\) Newtons

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