4
$\begingroup$

From the equation $E_k=\frac12mv^2$ you can determine more energy is necessary to accelerate a mass the higher your initial velocity is. For example, three times more energy is necessary to accelerate a mass from 1 m/s to 2 m/s than from 0 m/s to 1 m/s. If the earth is moving in the universe, what is our standard frame of reference? Why?

Thanks in advance for your time and answers

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ See physics.stackexchange.com/questions/535/… for understanding this... $\endgroup$
    – navigator
    Jan 26, 2016 at 13:55
  • $\begingroup$ The Earth is not moving "in the universe", but relative to other bodies in the universe. Kinetic energy is observer dependent and different observers would need different amounts of energy to accelerate the same body by the same amount relative to their inertial systems. $\endgroup$
    – CuriousOne
    Jan 26, 2016 at 16:29

1 Answer 1

Reset to default
0
$\begingroup$

The energy required to accelerate an object by a given velocity increment is linear in the initial velocity in the non-relativistic limit (where $E_k=\frac{1}{2}mv^2$ applies). It is even more energy intensive for the relativistic case when the velocity of light (c) is approached. That is because the relativistic expression for kinetic energy is: $$E_k=mc^2(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)$$.

Here is a site that should clarify both the relativistic and non-relativistic cases if you make use of the available links. http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html For your last question see the comments by @CuriousOne and @navigator above.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Using the work enrgy theorem it is aparent that it takes more energy to accelerate from 1 m/s to 2 m/s than form 0 m/s to 1 m/s. $\endgroup$
    – J Physics FM
    Jan 26, 2016 at 14:23
  • $\begingroup$ dE/dv=mv, i.e. it's a linear function of velocity, even in the classical case. $\endgroup$
    – CuriousOne
    Jan 26, 2016 at 16:25
  • $\begingroup$ @JPhysicsFM Must have been temporarily brain dead. I'll edit my answer. Proof that one should never rush a posting. $\endgroup$
    – Lewis Miller
    Jan 26, 2016 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.