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notes7a - kx W 2 2 1 2 2 1 − = − = θ θ x i x f x=0...

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Work done by a spring We know that work equals force times displacement But how to we calculate the work due to a non- constant force? Reconsider the restoring force of a spring It depends on the distance the spring is stretched or compressed. Hooke’s Law for the restoring force of an ideal spring. kx F s =
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x i x f x=0 d = x f x i But the force is not constant f f s i i s kx F kx F = = , , , Take the average force F s , avg = F s , i + F s , f 2 F s , avg = 1 2 k ( x f + x i ) Then the work done by the spring is W s = F s , avg cos φ d = F s , avg cos0 d = 1 2 k ( x f + x i )( x f x i ) = 1 2 k ( x f 2 x i 2 ) W s = 1 2 kx i 2 1 2 kx f 2
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2 2 1 kx U U s elastic = = Units of N/m m 2 = N m = J Total potential energy is 2 2 1 kx mgy U U U s g total + = + = Example Problem A block (m = 1.7 kg) and a spring (k = 310 N/m) are on a frictionless incline ( θ = 30°). The spring is compressed by x i = 0.31 m relative to its unstretched position at x = 0 and then released. What is the speed of the block when the spring is still compressed by x f = 0.14 m? f s i s f i s U U
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Unformatted text preview: kx W , , 2 2 1 2 2 1 − = − = θ θ x i x f x=0 x=0 Given: m=1.7 kg, k=310 N/m, θ =30°, x i =0.31 m, x f =0.14 m, frictionless Method: no friction, so we can use conservation of energy Initially θ sin , v v i 2 2 1 2 2 1 i i x h kx mgh m E = = + + = 2 2 1 sin i i i kx mgx E + = Finally 2 2 1 2 2 1 sin v v find , sin f f f f f f f kx mgx m E x h + + = = θ ) ( sin ) ( 2 v ) ( sin ) ( v sin sin v 2 2 2 2 2 1 2 2 1 2 2 1 2 2 1 2 2 1 f i f i f f i f i f i i f f f i f x x m k x x g x x k x x mg m kx mgx kx mgx m E E − + − = − + − = + = + + = s m 2 2 95 . 3 95 . 13 666 . 1 v ) 14 . 31 . ( 7 . 1 310 30 sin ) 14 . 31 . )( 8 . 9 ( 2 v = + = − + − = f f Interesting to plot the potential energies x f x f x f x i x i x i Energy U g U s U total E K...
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