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Springs

Springs. A coiled mechanical device that stores elastic potential energy by compression or elongation. Elastic Potential Energy – The energy stored in an object due to a deformation. Force Applied to a Spring. x 0 (unstretched position). F. x 1. F=k(x 1 -x 0 )

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Springs

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  1. Springs A coiled mechanical device that stores elastic potential energy by compression or elongation Elastic Potential Energy – The energy stored in an object due to a deformation.

  2. Force Applied to a Spring x0 (unstretched position) F x1 • F=k(x1-x0) • x0 is at the original unstretched position of the spring endpoint, x1= is the position of the spring after being stretched. Commonly written asF=kx, x=x1-x0 • x is the amount of compression or elongation from the equilibrium position measured in meters. • k – spring constant (force constant) – a measure of the resistance of the spring to deformation. • k is measured in N/m • The spring scales used in class have a spring constant of 350 N/m (a moderate spring constant)

  3. Restoring Force • The resistive force applied by the spring to return to its original shape is called the restoring force (Fs). • Fs=-kx. The restoring force is equal in magnitude and opposite in direction to the applied force. • F=-Fs • The restoring force applied by the spring is called Hooke’s Law. Hooke’s Law: Fs= -kx FS F

  4. Spring Constant Example • A 0.15 kg mass is attached to a vertical spring and hangs at rest a distance of 4.6 cm below its original position. An additional 0.50 kg mass is then suspended from the first mass and allowed to descend to a new equilibrium position. What is the total extension of the spring? • m1=0.15 kg x1=4.6 cm m2=0.50 kg x=stretched distance • k=F1/x1=(.15kg)(9.8m/s2)/.046m=32 N/m • x=F/k=(m1+m2)g/k • =(0.15 kg+0.50 kg)(9.8 m/s2)/(32 N/m)= • =0.20 m=20 cm x1 F1=m1g F=(m1+m2)g

  5. Work Accomplished on a Spring (Energy stored in a Spring) • Work is the area under a Force-Displacement graph. F W= ½ xF= ½ x(kx) The work accomplished in moving a spring from a zero reference to a position x. W= ½ kx2 x Since work is a transfer of energy, then potential energy is gained by a spring elongated from its reference position PEs= ½ kx2 Elastic potential energy gained by a spring.

  6. Work Moving a Spring between Two Locations The work (stored energy) moving the spring From position 1 to 2: W 12= W2-W1= ½ kx22- ½ kx12= ½ k(x22-x12) x1=0 for an unstretched spring W= ½ kx22 From position 2 to 3: W23= W3-W2= ½ kx32 - ½ kx22= ½ k(x32-x22) The work is moving the spring from position 1 to 2 or position 2 to 3 is NOT: W12= ½ k(x2-x1)2 W23= ½ k(x3-x2)2 Common mistake/misconception! x1 x2 x3

  7. Example of Work in Moving a Spring between Two Locations How much work is required to move a spring with a spring constant of 750 N/m from its unstretched position to 2.0 cm? W= W 01 =W1-W0 = ½ kx12- ½ kx02= ½ k(x12-x02)= ½ k(x12-0)= ½ kx12= ½ (750 N/m)(.02m)2=0.15 J How much work is required to move the same spring an additional 3.0cm? W=W 12 =W2-W1= ½ kx22- ½ kx12= ½ k(x22-x12)= ½ (750N/m)[(.05m)2-(.02m)2] =0.79 J

  8. Conservation of Mechanical Energy (Springs) v1 m Time 1 x1 x1=spring endpoint position at time 1 v1=mass velocity at time 1 x2=spring endpoint position at time 2 v2=mass velocity at time 2 v2 m Time 2 x2

  9. Springs Conservation of Mechanical Energy Equations • ME1=ME2 • PE1+KE1=PE2+KE2 • ½ kx12+ ½ mv12= ½ kx22+ ½ mv22

  10. Conservation of Mechanical Energy for SpringsExample Problem • A 10 kg mass traveling at 1.0 m/s on a frictionless surface compresses an originally undeformed spring with a constant of 800 N/m until it stops. What is the distance that the spring is compressed? • ME1=ME2 • PE1+KE1=PE2+KE2 • ½ kx12+ ½ mv12= ½ kx22+ ½ mv22 0+ ½ (10 kg)(1.0 m/s)2= ½ (800 N/m)x22+0 x2= 0.11 m = 11 cm

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