Flowers and physics. speed.

Question

A 4.5 kg flower pot sits precariously on a windowsill h1=31.9 meters above the ground. Your apartment is h2=14.6 meters above the ground, directly under the flower pot. If the flower pot falls past your window, what will be its speed?

Answer 1

Hi Claire!

 

Taking the windows to be "points" located at the given heights, we can use conservation of (mechanical) energy.  To do this, we can assess the value of mechanical energy at two points.  Mechanical energy (ME) is defined as:

 

ME = KE + PE

 

In words, it is the sum of the kinetic energy and all potential energies in the system.  We usually deal in this context only with the potential energy due to gravity, and that due to springs.

 

Potential energy due to gravity is given by:

PE_g = mgh

 

where h is the height above some "zero" level that we get to select.

 

Kinetic energy is given by:

 

KE = (1/2)mv²

 

where v is the speed (without regard to direction) of the object in question.

 

So, let's look at the mechanical energy at two different points.  The first is when the pot drops off the window sill, effectively being dropped, so that its initial speed is zero.  The mechanical energy as this point is:

 

ME_o = KE_o+(PE_g)_o 

 

Since the pot starts with zero velocity upon dropping off the upper sill, it starts with zero kinetic energy (KE_o = 0)

To determine its gravitational potential energy, we first have to choose a "zero" level.  The ground is often a reasonable choice.  But, in this case, it may be an advantage to set the "zero" level at the height of the lower window.  In this case, the height of the flower pot at the start of this motion is:

 

h_o = 31.9 m - 14.6 m = 17.3 m

 

Then, we can write the initial mechanical energy as:

 

ME_o = 0 + mgh_o

 

I'll leave that as is for now, without putting in numbers yet, for a reason we can see in a bit. 

 

Then, we can assess the mechanical energy as the flower pot is at the height of the lower window during its fall.  At this point:

 

ME_f = KE_f + (PE_g)_f

 

If the pot is at the height of the second window, which was defined as "zero" level above, then its gravitational potential energy at that point must be zero, because its final height, h_f, is 0 ((PE_g)_f = 0).

 

It is clearly moving with some speed at this point, which we will be solving for, so all of its mechanical energy is kinetic.

 

ME_f = KE_f + 0 = (1/2)mv_f²

 

Since only gravity is doing work here, mechanical energy is conserved, meaning that the initial and final mechanical energies are the same.

 

ME_o = ME_f

 

mgh_o = (1/2)mv_f²

 

Note that the mass, m, cancels out.  This is why I did not put in numbers yet.  In this idealized case, the mass of the falling object does not matter.  You may remember that this was also true in free fall (and all of kinematics):  mass of the object was never a concern.

 

v_f² = 2gh_o

 

So, we can now solve for the final speed, v_f, be some algebra using the equation above, and knowing h_o = 17.3 m.

 

I hope this helps!  Let me know if you have any questions about this or would like to check an answer.