What is the final speed of the block before it hits the ground in each case?

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To determine the final speed of the 1.7 kg block before it hits the ground, the work-energy principle is applied for two scenarios: a frictionless table and a table with kinetic friction. In the frictionless case, the conservation of energy indicates that the potential energy of the hanging mass converts entirely into kinetic energy, resulting in a specific speed calculation. For the case with a coefficient of kinetic friction of 0.15, friction must be accounted for, which reduces the speed of the block just before impact. Participants in the discussion emphasize the importance of understanding the conservation of energy equations for both scenarios to arrive at accurate results. The conversation highlights the need for clarity in applying these principles in physics problems.
ChrisMC
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In the figure, m1 = 3.0 kg. Use work and energy to find the speed of the m2 = 1.7 kg block just before it hits the floor in each of the following cases.

http://www.webassign.net/knight/p11-48alt.gif

(a) The table is frictionless.
. m/s
(b) The coefficient of kinetic friction of the 3.0 kg block is 0.15.
m/s


F= mg
F= mg-Muk*n
 
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Chris

Please show some attempt at a solution using the requested work-energy approach. I would suggest using the conservation of total energy principle for each case.
 
I honestly don't know where to start, I thought if the mass on top was heavier than the mass hanging wouldn't move.
 
ChrisMC said:
I honestly don't know where to start, I thought if the mass on top was heavier than the mass hanging wouldn't move.
well,no, on a frictionless surface, for example, the mass on the table could be enormous, but it would move with just the slightest force acting on it (the hanging mass could be therefore quite small). Are you familiar with the conservation of energy equations? Try writing them for the frictionless case, and then for the friction case.
 

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