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Aim: How can we calculate the energy of a spring?

Aim: How can we calculate the energy of a spring?. HW #33 due tomorrow. Do Now: An object is thrown straight up. At the maximum height, it has a potential energy of 300 J. How much work did it take to do this?. 300 J (Work – Energy Relationship).

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Aim: How can we calculate the energy of a spring?

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  1. Aim: How can we calculate the energy of a spring? HW #33 due tomorrow Do Now: An object is thrown straight up. At the maximum height, it has a potential energy of 300 J. How much work did it take to do this? 300 J (Work – Energy Relationship)

  2. Convert the mass into kg and calculate the weight: distance stretched (cm) mass (g)mass (kg)weight (N) 50 250 450 650 850 0.49 2.45 4.41 6.37 8.33 0.050 0.250 0.450 0.650 0.850

  3. Plot the data 10 9 8 7 6 5 4 3 2 1 0 Force or weight (N) Your graph should be a straight line similar to this one! Distance stretched (cm)

  4. Calculate the slope of your graph

  5. Hooke’s Law F = Force or weight applied (N) k = Spring constant (N/m) Each spring has a different value (slinky – low k) (shocks in car – high k) *show springs* x = Distance stretched or compressed from equilibrium (m) F = kx

  6. Robert Hooke 1635-1703

  7. Problems • A 4 N weight is hung from a spring causing it to stretch a distance of 0.16 m. What is the spring constant? F = kx 4 N = k(0.16 m) k = 25 N/m

  8. Next Problem • A 5 kg mass is hung from the same spring. How far does it stretch? F = kx mg = kx (5 kg)(9.8 m/s2) = (25 N/m)x x = 1.96 m

  9. Elastic Potential Energy • The energy stored in a spring or any other elastic item (i.e. – rubberband) • Represented as PEs • PEs = ½kx2 • On a F-x graph: Slinky Commercail Ace Ventura – Slinky Scene Area = PEs F x

  10. Problems • A spring with a constant of 30 N/m stretches 0.5 m. What is its elastic potential energy? • PEs = ½kx2 • PEs = ½(30 N/m)(0.5 m)2 • PEs = 3.75 J

  11. Next Problem 2. A spring with a constant of 40 N/m has 12 J of elastic potential energy. How far has it stretched? PEs = ½kx2 12 J = ½(40 N/m)x2 x = 0.77 m

  12. Solve for the elastic potential energy PEs = area under curve PEs = ½bh PEs = ½(2 m)(70 N) PEs = 70 J 70 Force or weight (N) 2 Distance stretched (m)

  13. Conservation of Energy An 80 kg girl jumps on a trampoline with 1000 J of elastic potential energy stored in the springs. How high with she go? PEs = PE PEs = mgh 1000 J = (80 kg)(9.8 m/s2)h h = 1.3 m

  14. Conservation of Energy A 100 kg object is connected to a spring with a spring constant of 5,000 N/m and is compressed a distance of 0.75 m. How fast will the object travel when released? PEs = KE ½kx2 = ½mv2 kx2 = mv2 (5,000 N/m)(0.75 m)2 = (100 kg)v2 v = 5.3 m/s

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