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Rollercoaster A 1700 kilogram rollercoaster operating on a frictionless track has a speed of 5 meters per second as it passes over the crest of a 35 meter.

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Presentation on theme: "Rollercoaster A 1700 kilogram rollercoaster operating on a frictionless track has a speed of 5 meters per second as it passes over the crest of a 35 meter."— Presentation transcript:

1 Rollercoaster A 1700 kilogram rollercoaster operating on a frictionless track has a speed of 5 meters per second as it passes over the crest of a 35 meter high hill. What is the rollercoaster’s PE at this point? What is the rollercoaster’s KE at this point? What will its total energy be at the bottom of the hill? What will its speed be at the bottom of the hill?

2 Aim: How do we account for the force of a spring?
HW:

3 Hooke’s Law To hold a spring either stretched or compressed an amount x from it’s natural (unstretched) length requires the hand to exert a force on the spring FP, that is directly proportional to x. That is, FP=kx Where k is a constant, called the spring stiffness constant, and is a measure of the stiffness of the particular spring. The stretched or compressed spring exerts a force Fs in the opposite direction on the hand. The force exerted on an object (F) is proportional to the amount of stretch or compression (x) and the spring constant (k) F = Newtons x = meters k = N/m

4 More Hooke’s Law Hooke’s Law mainly applies to elastic objects like springs and rubber bands We will mainly focus on springs All elastic objects obey Hooke’s Law until they reach their ELASTIC LIMIT and become deformed Equilibrium Position: The unstretched position.

5 Station Activity Go to your respected station and show all your work with the correct units provided.

6 EXAMPLES A spring with a spring constant of 10 newtons per meter is stretched 0.5 meters. How much force is applied to the spring? F = kx = (10 N/m)(0.5 m) = 5 N A spring with a spring constant of 15 newtons per meter is compressed with a force of 4.5 newtons. How far will the spring compress? F = kx 4.5 N = (15 N/m) x x = 0.3 m

7 EXAMPLES A spring is stretched a distance of 0.4 meters with a force of 20 newtons. Calculate the spring constant of the spring. F = kx 20 N = k (0.4 m) k = 50 N/m A 3000g mass is hung from the same spring above. How far does it stretch? (3 kg x 9.8 m/s2) = (50 N/m)x X = 0.59 m

8 Demo Lets find the spring constant for a spring really quick! If I hang the spring from a force probe sensor, we can measure the force vs. distance. What is my slope??? How much will a mass at rest stretch the spring, according to this slope? When setting up force probe, just plug into computer and correct graph should pop up automatically. Once k is known, the spring will stretch based on the weight of the mass divided by the constant “k”

9 Demo Cont. So now lets stretch the spring an additional 5 cm using a ruler! What is the force of the spring? With what speed will the mass reach the equilibrium point using kinematics? FNET = ma Vf2 = Vi2 + 2ad Lets prove this now, but how?

10 Why it no match? Acceleration is NOT constant! Kinematics will not work! We must use energy equations! Find PEs, then speed from KE formula!

11 SUMMARY 1. Which objects does Hooke’s Law apply to? Equation? 2. Which letter represents the spring constant? 3. Units for the spring constant? 4. Which kind of force acts on a spring if something is hung on the spring?

12 EXAMPLES A spring with a spring constant of 10 newtons per meter is stretched 0.5 meters. How much force is applied to the spring? A spring with a spring constant of 15 newtons per meter is compressed with a force of 4.5 newtons. How far will the spring compress?

13 EXAMPLES A spring is stretched a distance of 0.4 meters with a force of 20 newtons. Calculate the spring constant of the spring. A 3000g mass is hung from the same spring above. How far does it stretch?


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