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: mass (g)mass (kg)weight (N) 50 250 450 650 850 0.050 0.250 0.450 0.650 0.850 0.49 2.45 4.41 6.37 8.33 distance stretched (cm)

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

4. Calculate the slope of your graph

5. Hooke’s Law F = kx 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)

6. Robert Hooke 1635-1703

7. Problems 1.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 2.A 5 kg mass is hung from the same spring. How far does it stretch? F = kx mg = kx (5 kg)(9.8 m/s 2 ) = (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 PE s PE s = ½kx 2 On a F-x graph: F x Area = PE s Slinky Commercail Ace Ventura – Slinky Scene

10. Problems 1.A spring with a constant of 30 N/m stretches 0.5 m. What is its elastic potential energy? PE s = ½kx 2 PE s = ½(30 N/m)(0.5 m) 2 PE s = 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? PE s = ½kx 2 12 J = ½(40 N/m)x 2 x = 0.77 m

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

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? PE s = PE PE s = mgh 1000 J = (80 kg)(9.8 m/s 2 )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? PE s = KE ½kx 2 = ½mv 2 kx 2 = mv 2 (5,000 N/m)(0.75 m) 2 = (100 kg)v 2 v = 5.3 m/s