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# EA3_energy - EA3 Systems Dynamics IV Work and Energy Work...

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EA3: Systems Dynamics Work and Energy Sridhar Krishnaswamy 1 IV Work and Energy Newton’s laws provide us the tools necessary to track the motion of a body under the action of external forces. We have seen how to go about that for several situations, some of which were amenable to analytical solutions, and other cases could only be solved numerically. We will now explore some additional concepts of classical mechanics that provide us more insight into the workings of nature. IV.1 Work and Kinetic Energy: Let F be the net force exerted on a body by something external to it. We define the work U done by the force F in moving the body through some distance from point r i to point r f as: U = F . dr r i r f (4.1) where the integral is taken along the path of the motion. The units of work are evidently N.m (which is also known as the joule J) in SI units. Note that by definition (because of the dot product) it is only the component of the force that acts tangential to the path that contributes to the work. Next we define the kinetic energy of a body in motion as: K = 1 2 m v 2 (4.2) where m is the mass of the body and v is the instantaneous velocity. You can check that the dimensions of energy are the same as those of kinetic energy, and therefore the unit of kinetic energy is also the joule.

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EA3: Systems Dynamics Work and Energy Sridhar Krishnaswamy 2 Because of the force F acting on it, the body accelerates and so its velocity changes from say v i to v f .. According to Newton’s second law, we have: F = m d v dt (4.3) Let us now do some algebraic manipulation of the above. First, take the dot product of the above equation with v : F . v = m d v dt . v (4.4) Note that we can use the product rule of differentiation to get a useful identity: d dt ( v.v ) = d v dt . v + v . d v dt = 2 d v dt . v (4.5) and so (4.4) can be re-written as: F . d r dt = 1 2 m d dt v 2 ( 29 (4.6) where we have recast the velocity on the left side in terms of the rate of change of position, and we have used the identity (4.5) to recast the right side in terms of the magnitude of the velocity. Now, since (4.6) holds true over any time interval dt we must have: F . d r = 1 2 m d v 2 ( 29 (4.7) which upon integration from the initial point to the final point gives: F . d r r i r f = 1 2 m d v 2 ( 29 v i v f = 1 2 m v f 2 - v i 2 ( 29 (4.8) That is: U = ∆ K (4.9)
EA3: Systems Dynamics Work and Energy Sridhar Krishnaswamy 3 which reads: the work done by the net external force acting on a body is equal to the resulting change in the kinetic energy of the body. Remarks:

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EA3_energy - EA3 Systems Dynamics IV Work and Energy Work...

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