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

In summary, the conversation is about trying to solve a problem involving Kinetic Energy, specifically the height of a house when a stone falls from its roof. The formula for change in kinetic energy is discussed and it is mentioned that there are two unknowns and only one equation, so another way to relate the velocities is needed. One person suggests using motion under a constant force to solve the problem and provides a solution using this method.
  • #1
fara0815
45
0
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 !
 
Last edited:
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  • #2
Hi fara0815,

Your problem seems to be a misuse of the formula [tex] \Delta E_{kin} = \frac{1}{2}mv^2_f - \frac{1}{2}mv_i^2 [/tex]. This is not equal to [tex] \frac{1}{2} m v^2 [/tex] where [tex] v = \frac{v_f - v_i}{2} [/tex] is the average velocity. As it stands now, you have two unknowns [tex] v_i [/tex] and [tex] v_f [/tex] 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.
 
  • #3
Wow ! Physics Monkey, thank you very much!

I took this way:

[tex]
\Delta E_{kin} = \frac{1}{2}mv^2_f - \frac{1}{2}mv_i^2 [/tex]

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

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 !

[tex]
\Delta E_{kin} = \frac{1}{2}mv^2_f - \frac{1}{2}mv_i^2 [/tex]
 

1. What is kinetic energy?

Kinetic energy is the energy that an object possesses due to its motion. It is a type of energy that is associated with moving objects.

2. How is kinetic energy calculated?

Kinetic energy is calculated by using the equation KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity (speed).

3. What are some real-life examples of kinetic energy?

Some examples of kinetic energy in everyday life include a moving car, a spinning top, a thrown ball, and a running human.

4. Can kinetic energy be converted into other forms of energy?

Yes, according to the law of conservation of energy, kinetic energy can be converted into other forms such as potential energy, thermal energy, and sound energy.

5. How is kinetic energy related to potential energy?

Kinetic energy and potential energy are two forms of mechanical energy. Kinetic energy is the energy of motion, while potential energy is the energy that an object has due to its position or state. The two can be converted into each other. For example, a ball at the top of a hill has potential energy, but as it rolls down the hill, its potential energy is converted into kinetic energy.

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