Question

Hey can I get some insight on this? A man standing in an upwards-moving elevator drops a coin as the floor of the elevator passes the third floor (there’s a hole in the floor of the elevator so the coin drops down the elevator shaft).

Answer 1

elevators move at a constant velocity, so the coin will be experiencing 0 force on it becuase F = ma. see, if velocity is constant, then acceleration is zero, so now F = m*0. which in other words is saying the coin is essentially at rest, only it's position is increasing, but not it's acceleration. so as the man reaches the third floor, and drops the coin, it is the same as though he was standing still at the third floor and drops the coin.

Answer 2

would the coin have kinetic energy though?

Answer 3

yes, the coin will have kinetic energy, KE, becuase KE = (1/2)mv^2 and we are moving in an elevator, so there is the term v, or velocity.

Answer 4

so would the coin not hit the ground with more ke because it had inital ke plus the same gravitational energy as the other coin?

Answer 5

so if we are considering using kinetic and potential energy in this problem, the initial energy will look like this:
PE - KE = initial energy
mgh - (1/2)mv^2 = initial energy
the PE comes because we are 3 stories u
the -KE comes becuase we are going UP and elevator, the negative sign is there because velocity is moving in the opposite direction of the direction our coin will fall when dropped. if the elevator was going down, then kinetic energy would ADD TO the initail energy, but in our case it will take away.