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SO FAR WE HAVE DEALT WITH TWO KINDS OF POTENTIAL ENERGY: GRAVITATIONAL (U=MGH) ELASTIC (U=1/2KX 2 ) SOMETIMES IT IS MORE HELPFUL WHEN YOU HAVE POTENTIAL.

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Presentation on theme: "SO FAR WE HAVE DEALT WITH TWO KINDS OF POTENTIAL ENERGY: GRAVITATIONAL (U=MGH) ELASTIC (U=1/2KX 2 ) SOMETIMES IT IS MORE HELPFUL WHEN YOU HAVE POTENTIAL."— Presentation transcript:

1 SO FAR WE HAVE DEALT WITH TWO KINDS OF POTENTIAL ENERGY: GRAVITATIONAL (U=MGH) ELASTIC (U=1/2KX 2 ) SOMETIMES IT IS MORE HELPFUL WHEN YOU HAVE POTENTIAL ENERGY AS A FUNCTION OF POSITION TO FIND POTENTIAL ENERGY FIRST AND DERIVE THE FORCE FROM THERE. Potential Energy Curves

2 Force and Potential Energy We established last class that the relationship between work and potential energy was: which leads to…

3 Force and Potential Energy which leads to…. A conservative force always acts to push the system toward a lower potential energy.

4 A FORCE PARALLEL TO THE X-AXIS ACTS ON A PARTICLE MOVING ALONG THE X-AXIS. THE FORCE PRODUCES A POTENTIAL ENERGY: U(X) = 1.2X 4. WHAT IS THE FORCE WHEN THE PARTICLE IS AT X=-0.8M? Example

5 Force and Potential Energy This analysis can be extended to apply to three dimensions:

6 A PARTICLE MOVING ALONG THE X-AXIS IS ACTED ON BY A CONSERVATIVE FORCE. AT A CERTAIN POINT, THE FORCE IS ZERO. WHAT DOES THIS TELL YOU ABOUT THE VALUE OF THE POTENTIAL ENERGY FUNCTION AT THIS POINT? Check Point

7 Potential Energy Curves for a Spring Note: When the spring is either in a state of maximum extension or compression its potential energy is also a maximum When the spring's displacement is DOWN the restoring force is UP When the potential energy function has a negative slope, the restoring force is positive and vice-versa When the restoring force is zero, the potential energy is zero At any point in the cycle, the total energy is constant: U + K = U max = K max

8 Potential Energy Curve for a function

9 Points of Equilibrium When the force acting on the object is zero, the object is said to be in a state of EQUILIBRIUM! STABLE EQUILIBRIUM – located at minimums, if the object is displaced slightly it will tend back to this location. UNSTABLE EQUILIBRIUM – located at maximums, if the object is displaced slightly it will tend away from this location. STATIC EQUILIBRIUM – located at plateaus, where the net force equals zero.

10 Points of Equilibrium Stable Equilibrium at x 3 and x 5. Unstable Equilibrium at x 4. Static Equilibrium at x 1 and x 6.

11 Turning Points Define the boundaries of the particle’s motion. We know that E=K+U, so where U=E, K= 0 J and the particle changes direction. For instance, if E= 4J, there would be turning point at x 2..

12 If E = 1 J, why is the grey area referred to as an “energy well”? Turning Points

13 Example A particle of mass 0.5 kg obeys the potential energy function: U(x) = 2(x - 1) - (x - 2) 3 What is the value of U(0)? What are the values of x 1 and x 2 ?

14 Example A particle of mass 0.5 kg obeys the potential energy function: U(x) = 2(x - 1) - (x - 2) 3 How much potential energy does the particle have at position x 1 ? If the object was initially released from rest, how fast is it moving as it passes through position x 1 ?

15 Example A particle of mass 0.5 kg obeys the potential energy function: U(x) = 2(x - 1) - (x - 2) 3 How much potential energy does the mass have at x 2 ? How fast is it moving through position x 2 ?

16 Example A particle of mass 0.5 kg obeys the potential energy function: U(x) = 2(x - 1) - (x - 2) 3 Which position, x 1 or x 2, is a position of stable equilibrium?

17 Example A particle of mass 0.5 kg obeys the potential energy function: U(x) = 2(x - 1) - (x - 2) 3 How fast is the particle moving when its potential energy, U(x) = 0 J? If x 3 = ½x 1, then how fast is the particle moving as it passes through position x 3 ?

18 Example A particle of mass 0.5 kg obeys the potential energy function: U(x) = 2(x - 1) - (x - 2) 3 Sketch the graph of the particle's acceleration as a function of x. Indicate positions x 1 and x 2 on your graph.

19 Example A particle of mass 0.5 kg obeys the potential energy function: U(x) = 2(x - 1) - (x - 2) 3 At what value of x does the particle experience it greatest negative acceleration? What is the value of its potential energy at this position? How much kinetic energy does it have at this position? What force is being exerted upon it at this position? What is the value of its acceleration at this position?


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