vikram_gupta11 Posts: 44, Reputation: 1 Junior Member #1 Apr 30, 2017, 06:44 AM
Velocity transfer
Two 100 kilogram balls are mounted on a seesaw and this seesaw is in balanced position .Now I want to know that how much force I should apply on the right side arm of this balanced seesaw so that the left side 100 kilogram weight ball could jump in the air up to 5 centimeter.

This question is regarding a project and it is not a homework related.
 ebaines Posts: 12,130, Reputation: 1307 Expert #2 Apr 30, 2017, 08:41 AM
I assume what you mean is you want the 100 Kg ball on the left to fly up off the seesaw when the right side hits the ground and stops - correct? You'll need to tell us the travel distance over which the force on the right side operates - in other words how far does the right side travel from the startng equilibrium point before hitting the ground?
 vikram_gupta11 Posts: 44, Reputation: 1 Junior Member #3 Apr 30, 2017, 11:16 AM
The travel distance of right side will be 10 centimeter from the ground.
 ebaines Posts: 12,130, Reputation: 1307 Expert #4 Apr 30, 2017, 11:50 AM
You will need about 1000 N force applied downward on the right hand side of the seesaw.
 vikram_gupta11 Posts: 44, Reputation: 1 Junior Member #5 Apr 30, 2017, 12:15 PM
 ebaines Posts: 12,130, Reputation: 1307 Expert #6 May 1, 2017, 05:33 AM
Use energy principles:

1. The velocity you want at the instant the seesaw stops is v^2 = 2gh, where h = 0.05 meters (the height the 100 Kg mass needs to rise above the stopping point).

2. The work put into the system is F times d, where F is the applied force and d is the distance the force is applied over, which is 0.1m.

3. Conservation of energy:
Change in kinetic energy of the two masses + change in potential energy of the two masses = work applied.

Can you take it from here?

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