Kinetic Energy: It sounds simple but I have been having a very hard time

[fara0815]

Hello there !
I have been trying to figure the following problem out but unfortunately I have not had any obvious success. It is about Kinetic Energy:

A stone is falling from the roof of a house down to the ground. In the last second of the fall, the stone's energy increases of 293 J.
How high is the house?
(friction is neglected)

I have gone several ways but never got the stated result of 20,06 m.
I would be very happy if anyone could give me a hint to solve this one. It seems so easy but I cannot make it :(

The most logical way - at least for me - would be: (the ground is zero in my coordinate system)

∆Ekin= 293 J = 1/2 * m * v^2 thus: V= - 17,13 m/s

V is the velocity on average in the last second. Thus the calculated velocity is actually there at tf-0,5s. With Tf is meant the over all time of the fall.

V= - g*t, thus t= 1.745 s
tf= t + 0.5 s = 2,245 s

h= 1/2*g*t^2= 24,72 m

Any suggestions ?
I would be really happy to get some help with this one !

 

[Physics Monkey]

Hi fara0815,

Your problem seems to be a misuse of the formula \Delta E_{kin} = \frac{1}{2}mv^2_f - \frac{1}{2}mv_i^2. This is not equal to \frac{1}{2} m v^2 where v = \frac{v_f - v_i}{2} is the average velocity. As it stands now, you have two unknowns v_i and v_f and only one equation relating them (the change in kinetic energy equation). You need to find some other way to relate the velocities, but this is simple to do since you're looking at motion under a constant force.

 

[fara0815]

Wow ! Physics Monkey, thank you very much!

I took this way:

<br /> \Delta E_{kin} = \frac{1}{2}mv^2_f - \frac{1}{2}mv_i^2

= \frac{1}{2}ma^2 t^2_f - frac{1}{2}ma^2 (t_f - 1s)^2
= ma^2 t_f - \frac{1}{2}ma^2

Thus t_f = 2,022 s and with that is h= 20,06 m !


I just found this forum and I will definitely take a look around if I can help someone here! The Internet is such a great thing !

<br /> \Delta E_{kin} = \frac{1}{2}mv^2_f - \frac{1}{2}mv_i^2