Kinetic and Potential Energy Relationship Explained - Solve for One-Third KE

  • Thread starter Thread starter rkslperez04
  • Start date Start date
  • Tags Tags
    Potential
AI Thread Summary
The discussion focuses on the relationship between kinetic and potential energy, emphasizing the conservation of mechanical energy, which states that the total mechanical energy remains constant in the absence of external forces like friction. A specific problem is presented where a ball is dropped from a height of 60 meters, and the goal is to determine the height at which one-third of its total energy is kinetic energy. The calculations show that the total energy at the initial height is 588m, leading to a kinetic energy of 196m when one-third of the total energy is reached. By applying the conservation of energy equation, it is determined that the height above the ground when one-third of the energy is kinetic is 40 meters. The discussion also clarifies that when a ball is dropped, its initial kinetic energy is zero, but it gains kinetic energy as it falls, while potential energy decreases.
rkslperez04
Messages
31
Reaction score
0
I was in class today and we flew through the relantionship with Kinetic engery and Potential engery rather quickly.

I looked it up in the book but it went over my head.

Ei = Ef

Meaning
Ei = Ef
^ ^
Kinetic intital + Potential intital = Kinetic Final + Potential Final


This relates to the following question:

A ball is dropped from a hieght of 60meters. How high above the ground will it be when one-third of its total engery is KE?


Can you tell me the the Ei = Ef means and then give me a starting point to work the problem?
 
Physics news on Phys.org
Kinetic intital + Potential intital = Kinetic Final + Potential Final

That is correct. This an example of conservation of energy, and is known as conservation of mechanical energy (Kinetic and potential). It says, the mechanical energy at any time is a constant, in the absence of forces such as friction.
 
i think
the energy is initial potential energy at 60m and kinetic energy is zero
and as the ball is dropped
it comes down decreesing the potential energy and increasing the kinetic energy and at the utmost bottom, there is maximum kinetic energy
so kinetic energy increases coming downward and potential energy
at 60 meter, total Energy= PE= m*g*h= 9.8*60*h=588m
as illustrated KE= total energy/3 = 588m/3 = 196m
now placing in equation
TE= KE + PE or PE1+KE1 = PE2+KE2,
where KE1=0, PE1=588m, KE2= 196m and PE2=?
0+588m=196m+mgh2 suppose reuired height is h2
cutting m from all equation
588=196+gh2 => gh2=588-196 => h2= 392/g = 392/9.8 = 40
so the height will be 40 meters
 
ohh ok... let me rework the problem now...

soo when you drop a ball... the KE on the ball is 0 being its like Intial Velocity.. no movement yet... but as it goes... it gains kinetic engery ??
 
yup u got it. coz initially the ball is released instead of giving it a force for it to move.as it moves,the ball lose PE while gaining KE meaning energy is conserved. there is another point i want to add...when u throw a ball vertically upwards,initial velocity is not zero...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top