Conservative forces and the potential energy function. Nonconservative forces and the workenergy theorem


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1 Nonconservative forces and the workenergy theorem Consider an object falling with airresistance. There are two forces to consider; the gravitational force (conservative) and the drag force (nonconservative). 1 The total work done is W 1! = W c + W nc, where W c is the work done by the conservative (gravitational) force and W nc is the work done by the nonconservative (drag) force. But, by the workenergy theorem W 1! = "K = K # K 1. Also W c = #"U = #( U # U 1 ). ( ) + W nc, ( ) # K 1 + U 1 $K # K 1 = # U # U 1 i.e., W nc = K + U ( ), or W nc = E mech # E mech1 = "E mech. Therefore, the work done by a nonconservative force is equal to the change in mechanical energy. Conservative forces and the potential energy function If the potential energy of a system has a unique value at every point! r, we can define a potential energy function, U(! r ), which tells us how potential energy varies with position. One property of a conservative force is that the work done by the force can be epressed as the difference between the initial and final values of the potential energy, i.e., dw =! F d! s =!du, where U is the potential energy function. Hence, the work done by a conservative force equals the decrease in the potential energy from point 1 to point. "#U = U! U 1 =! $ F! d! s. Since du =!! F d! s, then, in onedimension, du =!F.d, i.e., F =! du d. 1
2 Given that F =! du, if we know the potential energy d function, U(), we can determine the force F at any point. Conversely, if we know the functional form of F, we can find the potential energy function, since Eamples... [1] Gravitational force: ˆ k m! g du =!F.d, i.e., U =!" F.d. Let us find the potential energy function for the gravitational force. We take the zdirection ( k ˆ ) vertical. Then, from above, the incremental change in potential energy as the object falls is du =! F! d! s =!(!mgk ˆ ) (dˆ i + dyˆ j + dzk ˆ ) = ( mg)dz. #U = mg dz " = mgz + U ", where U " is the reference energy when z = 0. Let s plot this function: U(z) 1 #U = U " U 1 U! Height (z) Note: U! U(z), i.e., it has a unique value at z. This is the potential energy function for the gravitational force. Normally, we deal only with differences in potential energy, so the choice of U! is entirely arbitrary. Note: the force associated with this potential energy function is F z = " du = "mg, as dz epected. [] Elastic force: F 1 = k F = "k If you compress a spring a distance in the  direction, the force, F 1, you eert is given by Hooke s Law, i.e., F 1 = k, where k is the spring constant. The spring does negative work since the force it eerts on you ( F = "k) is opposite to the displacement.
3 Therefore, the elastic potential energy function U() is given by: du =! F! d! s =!(!F ).d = k.d. "U() = k#.d = 1 k + U ". But, when there is no displacement, i.e., = 0, the spring U() Equilibrium position:!. Stable equilibrium when U(! ) is a minimum, i.e., d U d > 0. has zero elastic potential energy.! "U(0) = 0, so U " = 0. U() Equilibrium position:!. At this point, there is no net force acting on the spring (in Unstable equilibrium when the direction); we say that the spring is in equilibrium. U(! ) is a maimum, This is the definition of equilibrium, which we first came across in chapter 4 (Newton s 1st Law). i.e., d U d < 0. Equilibrium: At equilibrium F = 0, i.e., du d = 0, so U() is an etremum. However, we must eercise care in using that criterion, since du = 0 does not uniquely d U()! Equilibrium position:!. Neutral equilibrium since U(! ) is an infleion point, i.e., d U d = 0. define a minimum in the potential energy function.!
4 Question 10: The potential energy of an object, confined to move along the ais, is U() = 3! 3 for " 3 m, and U() = 0 for > 3 m, where U() is in Joules and is in meters. If the only force acting on the object is the force associated with this potential energy function, (a) at what positions is the object in equilibrium and what type of equilibrium is associated with each position? (b) Sketch the potential energy function. (a) To find the equilibrium positions we need F(). F() =! du() d =! F() (m)!1 1 At equilibrium F() = 0, i.e., du() = 6! 6 d = 6(!1) = 0. Therefore, the equilibrium positions are at = 0 and = 1 m. To find out the type of equilibrium look at the signs of the seond derivative, d U() d, at = 0 and = 1 m. Now, d U() d At = 0: d U() d = 6!1. = 6 (> 0) " Stable. At = 1 m: d U() d =!6 (< 0) " Unstable.
5 (b) U(J) 10 U() = 3! 3!1 1 (m)!10
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