derive the expression for kinetic energy.
Answers
Suppose a body of mass m is moving with velocity v. It is brought to rest by applying a retarding force F. Suppose it traverses a distance s before coming to rest. Kinetic energy of body, KE = Work done by retarding force to stop it.
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Answer:
∆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
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