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I have a 100 kg weight dropping 20 meters at 10 m/s^2

I calculate the Force of the falling weight as 100kg × 10m/s^2 = 1000 Newtons, and the Potential energy mgh = 1000 Newtons × 20 meters = 20,000 Joules, converted into Kinetic Energy 1/2 mv^2 when the weight drops

Time of drop as t = squareroot(2×20m/10m/s^2) = 2 seconds

And it's velocity as 2 s × 10m/s^2 = 20 m/s

Checking energy conservation: mgh = 1/2 mv^2 = 100kg/2×(20m/s)^2 = 20,000 Joules

I now attach the 100 kg weight to another 100 kg weight on a frictionless horizontal surface using a rope and single pulley wheel.

The acceleration of the attached weights = 100 kg/200 kg × 10m/s^2 = 5 m/s^2

Time of drop t = squareroot(2x20m/5m/s^2) = 2.83 seconds Velocity as 2.83 s × 5m/s^2 = 14.15 m/s for both objects

Checking energy conservation: 1/2mv^2 = 200kg/2×(14.15m/s)^2 = 20,000 Joules

Using Joules to directly calculate Velocity (as shown by @trula and Marco Ocram):

20,000 Joules = 1/2 × (100 kg (v m/s)^2 + 100 kg (v m/s)^2) = 1/2 × 200 kg v^2 m^2/s^2

v^2 = 200 m^2/s^2,
v = 14.14 m/s

I now attach a 1 to 3 gear to the horizontal weight, giving the falling weight a 3x mechanical disadvantage. It now moves the horizontal weight 3x as far (60 meters) when it drops.

Velocity of the drop weight:

20,000 Joules = 1/2 × (100kg × (v m/s)^2 + 100kg × (3v m/s)^2) = 1/2 × (100 v^2 + 900v^2) = 500 v^2 m^2/s^2

v^2 = square root (40 m^2/s^2) v = 6.32 m/s

Velocity of the 1:3 Geared horizontal object (which moves 3x faster and 3x the distance): 6.32 m/s × 3 = 18.96 m/s

Are these energy calculations consistent with conservation of energy theory?

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  • $\begingroup$ To get help with this you need to ask a specific conceptual question, as described here. Direct "check my work" style questions are off-topic, as explained here. $\endgroup$ – PM 2Ring Nov 6 '19 at 5:45
  • $\begingroup$ your last calculation does not agree with energy conservation . mgh=m/2*(v1^2+9v1^2) $\endgroup$ – trula Nov 6 '19 at 6:36
  • $\begingroup$ @trula May be correct with edits, thanks! $\endgroup$ – Robert DiGiovanni Nov 6 '19 at 17:00
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You can check for yourself by calculating the combined KE of the two weights given the speeds at which you concluded they were moving. If you do that you will find their combined KE is out by a factor of two, roughly. Where you have gone wrong is the equation under the heading 'Velocity of the drop weight', where you have double counted the weights.

Specifically, the KE of the suspended weight, 0.5*100*v2, plus the KE of the geared weight, 0.5*100*(3v)2 is 0.5*100*10v2, whereas you have equated it to 0.5*200*10v2.

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  • $\begingroup$ Fantastic Marco, thanks! $\endgroup$ – Robert DiGiovanni Nov 6 '19 at 16:20
  • $\begingroup$ I was trying to treat the connected drop weight and the connected horizontal weight as one object, but, obviously, I cannot with gears. That's why they call it engine, transmission, wheels (LOL). $\endgroup$ – Robert DiGiovanni Nov 6 '19 at 16:28

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