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Unformatted text preview: Phys 2101 Gabriela Gonzlez There is potential energy associated with the work of conservative forces (gravity: U=mgh; springs: U= kx 2 ): U = U f U i = W If the only forces doing work are conservative the sum of kinetic and potential energy is conserved: U + K = 0 U f + K f = U i + K If there are nonconservative forces, the change in the total energy (potential + kinetic) is equal to the work done by those forces: U + K = W other 2 3 U(h)=mgh h x U(x) = kx 2 Where is F=0? (equilibrium position) 4 U = W = F ( x ) dx F ( x ) = dU ( x ) dx Force = slope of U(x) curve Equilibrium: F=0 = dU/dx. d 2 U/dx 2 >0: stable eq; If E=KE+U, when U(x)=E, KE =0 and v=0: turning point If E=KE+U, since KE is +ve, U<E: only certain positions are allowed. 4 U = W = F ( x ) dx F ( x ) = dU ( x ) dx Force = slope of U(x) curve Equilibrium: F=0 = dU/dx. d 2 U/dx 2 >0: stable eq; If E=KE+U, when U(x)=E, KE =0 and v=0: turning point If E=KE+U, since KE is +ve, U<E: only certain positions are allowed. 5 Figure 848 shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) The particle is released at x = 4.5 m with an initial speed of 7.0 m/s, headed in the negative x direction. (a)If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x = 4.0 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 7.0 m/s. (d) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of x = 5.0 m?...
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 Fall '07
 GROUPTEST
 Physics, Energy, Force, Gravity, Potential Energy, Work

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