How Fast Will the Object Be Falling When It Passes Ken's Window?

[mt05]

hey everyone! I think this question can probably be solved using a constant acceleration formula, but since we are doing the momentum and work units in class i figured i was suppose to use those formulas to solve my question. anyways here it is :

Paul lives on the sixth floor of an apartment complex. His window is 20.2 m above the ground. Paul notices a 7.25 kg object falling past his window at 8.50 m/s. If Ken's window is 5.00m above the ground level, how fast will that same object be falling as it passes by Ken's window?

KE = 1/2(7.25)(8.50)^2
KE = 261.91
KE = W, W = FD, FD = 1/2mv^2final - 1/2mv^2initial

261.91 = 1/2(7.25)(v^2) - 1/2(7.25)(8.50)^2
261.91 = 3.625v^2 - 261.91
523.82/3.625 = v^2
vfinal = 12 m/s.
i know that is wrong, because when i put the numbers into the formula vf^2 = vi^2 + 2ad i get 19.2 m/s which seems more reasonble. Is there anyway to figure this question out properly by using momentum/work ?
KE = W, W = FD, FD = 1/2mv^2final - 1/2mv^2initial ( this doesn't seem to make much sense either, but i thought i saw it on my formula sheet )
thanks for any help!

 

[Redbelly98]

Welcome to PF

KE = 1/2(7.25)(8.50)^2
KE = 261.91
KE = W, W = FD, FD = 1/2mv^2final - 1/2mv^2initial

261.91 = 1/2(7.25)(v^2) - 1/2(7.25)(8.50)^2
261.91 = 3.625v^2 - 261.91

Work W is not equal to the initial KE, so it's wrong to simply plug in the initial KE of 262 J for W.

Instead, calculate what W=FD is by figuring out the values of F and D.

 

[kerimek]

Possible response:

Hi there! It's great that you are trying to use different formulas to solve this question. However, it seems like you may be mixing up the concepts of kinetic energy and work. Kinetic energy is the energy an object possesses due to its motion, while work is the transfer of energy from one object to another. In this case, we are dealing with the motion of the falling object, so using the formula KE = 1/2mv^2 is the most appropriate approach.

To solve this question, we can use the conservation of energy principle, which states that the total energy of a system remains constant. In this case, the system is the falling object. So, at Paul's window, the object has a kinetic energy of 261.91 J. As it falls and passes by Ken's window, its potential energy decreases and its kinetic energy increases, but the total energy remains the same.

Therefore, we can set up the equation:

KE at Paul's window = KE at Ken's window

1/2(7.25)(8.50)^2 = 1/2(7.25)(vfinal)^2

Solving for vfinal, we get vfinal = 8.50 m/s.

I hope this helps clarify the concept of energy and how to approach this type of question. Keep up the good work!