Derivative of energy is force?

[physicsisfeyn]

Why is the derivative of energy force? For example: KE=1/2mv^2 the derivative of which, assuming mass is held constant is KE'= ma. Why is this? What is the significance of this? What are some applications of this?

Additionally, Einstein's famous equation, E=mc^2, relates mass with energy. Does this mean as you approach the speed of light, that force is actually converted into mass? For example, if you have a mass of 1kg traveling with an acceleration of 299,000,000m/s and you double the force (which is 299,000,000N times 2) that some of the force is actually transformed into mass?

 

[tiny-tim]

welcome to pf!

hi physicsisfeyn! welcome to pf!

(try using the X² icon just above the Reply box )

Why is the derivative of energy force? For example: KE=1/2mv^2 the derivative of which, assuming mass is held constant is KE'= ma.

nooo … derivative of momentum is force: d(mv)/dt = ma

d(1/2 mv²)/dt = mv dv/dt = mva

if you mean d(1/2 mv²)/dv, that's momentum, mv (and the meaning is that the laws of physics are invariant under a change of velocity of the observer, so differentiating the conservation of energy wrt v must still give a conservation equation, and the only one left is momentum!)

i don't see any way of getting ma from 1/2 mv²

Additionally, Einstein's famous equation, E=mc^2, relates mass with energy. Does this mean as you approach the speed of light, that force is actually converted into mass?

no, e = mc² has nothing to do with the speed of light, the c² has no physical significance: it is simply the conversion factor between J and kg

if energy was measured in light-kilograms (in the same way that distance can be measured in light-years), then the conversion factor would be unnecessary, and the formula would be e = m

 

[Delta2]

Derivative of Energy or Work with respect to displacement s yields force. This is from the definition of work as integral of force over distance s and the basic theorem of calculus.

In the case of kinetic energy taking direvative wrt to s and have in mind that v=ds/dt we have dE/ds=mvdv/ds=mv(dv/dt)(dt/ds)=mavdt/ds=ma.

 

[NEILS BOHR]

F = dU/dx

and F = dP / dt

 

[physicsisfeyn]

Okay.

So am I wrong in thinking that mass increases with speed? If so, what happens when you approach the speed of light? If you're right at the barrier, say .9999c, and you double the force, what happens?

 

[tiny-tim]

you can keep increasing the momentum indefinitely

the momentum is m₀v/√(1 - v²/c²), which "approaches infinity" as v approaches c …

loosely speaking, doubling the momentum gets you twice as close to c

 

[granpa]

force times distance = energy

 

[AlephZero]

i don't see any way of getting ma from 1/2 mv²

As somebody else said the derivative of energy with respect to distance is force

Keep it simple and assume the mass stays constant:

d/dx (m v^2 / 2)
= m/2 d/dx (dx/dt)^2
= m/2 dt/dx d/dt (dx/dt)^2
= m/2 dt/dx (2 dx/dt) d^2x/dt^2
= m d^2x/dt^2
= ma

For example the energy stored in a spring = 1/2 K x^2
and the force in the spring = d/dx (1/2 K x^2) = Kx