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The Spring Consider a spring, which we apply a force FA to either stretch it or compress it FA x unstretched -FA -x x=0 FA kx k is the spring constant, units of N/m, different for different materials, number of coils From Newton’s 3rd Law, the spring exerts a force that is equal in magnitude, but opposite in direction Hooke’s Law for the restoring force of an ideal spring. It is s a conservative force. F kx Work and Potential Energy of a Spring We know that work equals (force times displacement) s x f xi But the force is not constant xi xf Fs ,i kxi , Fs , f kx f Take the average force Fs x=0 Fs ,i Fs , f 2 Fs 12 k ( x f xi ) Then the work done by the spring is Ws Fs cos s Fs cos 0 s 2 2 1 1 2 k ( x f xi )( x f xi ) 2 k ( x f xi ) 2 2 1 1 Ws 2 kxi 2 kx f U s ,i U s , f U elastic U s kx 1 2 2 Units of N/m m2 =Nm=J Total potential energy is Utotal U g U s mgh kx 1 2 2 Example Problem A block (m = 1.7 kg) and a spring (k = 310 N/m) are on a frictionless incline ( = 30°). The spring is compressed by xi = 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 xf = 0.14 m? x=0 xi x=0 xf Given: m=1.7 kg, k=310 N/m, =30°, xi=0.31 m, xf=0.14 m, frictionless Method: no friction, so we can use conservation of energy Initially E mv mgh kx vi 0, hi xi sin 1 2 2 1 2 Ei mgxi sin kx 1 2 2 i 2 Finally h f x f sin , find v f 2 2 1 1 E f 2 mv f mgx f sin 2 kx f E f Ei 2 2 1 1 2 mv f mgx f sin 2 kx f 2 1 mgxi sin 2 kxi 2 2 2 1 1 2 mv f mg ( xi x f ) sin 2 k ( xi x f ) k 2 2 v f 2 g ( xi x f ) sin ( xi x f ) m 310 v f 2(9.8)(0.31 0.14) sin 30 (0.312 0.14 2 ) 1.7 v f 1.666 13.95 3.95 ms Interesting to plot the potential energies Energy K Ug xf E Utotal Us xi xf xi xf xi