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### Kinetic energy

Posted: Fri Nov 02, 2018 4:53 pm
Does the formula (1/2)mv^2 only apply to kinetic energy? or can it be used for other types of energy?

### Re: Kinetic energy

Posted: Fri Nov 02, 2018 4:55 pm
That formula is specifically for kinetic energy so don't think it can be used for other types of energy.

### Re: Kinetic energy

Posted: Sat Nov 03, 2018 8:37 pm
As stated before the Formula is used only to find the Kinetic energy but if you don’t have all the requested information I’m pretty sure you can also use de Broglie equation to help you find that portion. Since the equation is KE= 1/2 MV^2 and you don’t have velocity you can use de Broglie’s equation to help you find that information if they give you lets say the mass and wave length. I hope this helps a bit.

### Re: Kinetic energy

Posted: Sat Nov 03, 2018 11:34 pm
It is only used for kinetic energy. It is specifically given as the equation for kinetic energy on the equations sheet so there is no reason to think that it is used for other types of energy

### Re: Kinetic energy

Posted: Sat Nov 03, 2018 11:37 pm
I agree. In our context it is frequently used in electron kinetic energy calculations and for relating debroglie principles to the photo electric effect.

### Re: Kinetic energy

Posted: Sat Nov 03, 2018 11:39 pm
We use it to only find kinetic energy!

### Re: Kinetic energy

Posted: Thu Dec 06, 2018 6:30 pm
Madeline Lequang 1G wrote:Does the formula (1/2)mv^2 only apply to kinetic energy? or can it be used for other types of energy?

It only applies to kinetic energy.

### Re: Kinetic energy

Posted: Sun Dec 09, 2018 7:12 pm
It only applies to kinetic energy. For example the energy of a photon is measured by plank's constant times it's frequency.

### Re: Kinetic energy

Posted: Sun Dec 09, 2018 7:21 pm
The formula only works for kinetic energy because kinetic energy is the only energy that is directly proportional to the mass of the object and to the square of its velocity

### Re: Kinetic energy

Posted: Sat Jun 29, 2019 12:13 pm
Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1.

What is the kinetic energy of the ejected electron?
A. 3.01 x 1025 J

B. 3.98 x 10-19 J

C. 7.96 x 10-19 J

D. 1.99 x 10-19 J

E. None of the above

How would you go about solving this problem? I know you would use the Ek=1/2mv^2 formula, but I don't know what the mass would be. Someone please help.

### Re: Kinetic energy

Posted: Sat Jun 29, 2019 1:40 pm
Nevermind, I figured it out. It's the mass of the electron, right?

### Re: Kinetic energy

Posted: Tue Jul 02, 2019 11:12 am
In my opinion, this formula is only for the kinetic energy, or sometimes the electron doesn't have kinetic energy.