write what is kinetic energy and derive equation 2 of kinetic energy
Answers
Answer:
kinetic energy is the energy
Explanation:
In physics, the kinetic energy of an object is the energy that it possesses due to its motion.[1] It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.
formula
Kinetic energy is a simple concept with a simple equation that is simple to derive. Let's do it twice.
Derivation using algebra alone (and assuming acceleration is constant). Start from the work-energy theorem, then add in Newton's second law of motion.
∆K = W = F∆s = ma∆s
Take the the appropriate equation from kinematics and rearrange it a bit.
v2 = v02 + 2a∆s
a∆s = v2 − v02
2
Combine the two expressions.
∆K = m ⎛
⎜
⎝ v2 − v02 ⎞
⎟
⎠
2
And now something a bit unusual. Expand.
∆K = 1 mv2 − 1 mv02
2 2
If kinetic energy is the energy of motion then, naturally, the kinetic energy of an object at rest should be zero. Therefore, we don't need the second term and an object's kinetic energy is just…
K = ½mv2
Derivation using calculus (but now we don't need to assume anything about the acceleration). Again, start from the work-energy theorem and add in Newton's second law of motion (the calculus version).
∆K = W
∆K = ⌠
⌡ F(r) · dr
∆K = ⌠
⌡ ma · dr
∆K = m ⌠
⌡ dv · dr
dt
Rearrange the differential terms to get the integral and the function into agreement.
∆K = m ⌠
⌡ dv · dr
dt
∆K = m ⌠
⌡ dr · dv
dt
∆K = m ⌠
⌡ v · dv
The integral of which is quite simple to evaluate over the limits initial speed (v) to final speed (v0).
∆K = 1 mv2 − 1 mv02
2 2
Naturally, the kinetic energy of an object at rest should be zero. Thus an object's kinetic energy is defined mathematically by the following equation…
K = ½mv2
Thomas Young (1773–1829) derived a similar formula in 1807, although he neglected to add the ½ to the front and he didn't use the words mass and weight with the same precision we do nowadays. He was also the first to use the word energy with its current meaning in a lecture on collisions given before the Royal Institution.
