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Lecture 13 Potential Energy

Lecture 13 Potential Energy. There are two types of forces: conservative (gravity, spring force) All microscopic forces are conservative: Gravity, Electro-Magnetism, Weak Nuclear Force, Strong Nuclear Force nonconservative (friction, tension) Macroscopic forces are non-conservative,.

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Lecture 13 Potential Energy

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  1. Lecture 13 Potential Energy

  2. There are two types of forces: • conservative (gravity, spring force) • All microscopic forces are conservative: • Gravity, • Electro-Magnetism, • Weak Nuclear Force, • Strong Nuclear Force • nonconservative (friction, tension) • Macroscopic forces are non-conservative,

  3. Conservative Forces A force is conservative if the work it does on an object moving between two points is independent of the path taken.  work done depends only on ri and rf

  4.  If an object moves in a closed path (ri = rf) then total work done by the force is zero.

  5. Nonconservative Forces  work done by the force depends on the path  non-conservative forces dissipate energy

  6. Work Done by Conservative Forces Potential Energy:Energy associated with the position of an object. For example: When you lift a ball a distance y, gravity does negative work on the ball. This work can be recovered as kinetic energy if we let the ball fall. The energy that was “stored” in the ball is potential energy. Wc = -DU =-[Ufinal – Uinitial] Wc = work done by a conservative force DU = change in potential energy

  7. Gravitational Potential Energy • Gravitational potential energy U = mg(y-y0),where, y = height • U=0 at y=y0 (e.g. surface of earth). • Work done by gravity: • Wg = -mg Dy = -mg (y- y0)

  8. Spring Potential Energy Uf – Ui = - [Work done by spring on mass] Mass m starts at x=0 (Ui =0) and moves until spring is stretched to position x. WorkSpring = - ½ kx2 U(x) – 0 = - (-1/2 kx2) USpring(x) = ½ kx2 x = displacement from equilibrium position F=-kx x Area in triangle = - kx times increment in x = Work done by spring

  9. Conservation of Energy • Energy is neither created nor destroyed • The energy of an isolated system of objects remains constant.

  10. Mechanical Energy (Conservative Forces) Mechanical energyE is the sum of the potential and kinetic energies of an object. E = U + K The total mechanical energy in any isolated system of objects remains constantif the objects interact only through conservative forces: E = constant Ef =Ei Uf + Kf = Ui+ Ki DU + DK = DE = 0

  11. Example: A ball of mass m is dropped from a height h above the ground, as shown in Figure (A) Neglecting air resistance, determine the speed of the ball when it is at a height y above the ground. (B) Neglecting air resistance, determine the speed of the ball when it is at the ground.

  12. Example:  A 0.5 kg block is used to compresses a spring with a spring constant of 80.0 N/m a distance of 2.0 cm.  After the spring is released, what is the final speed of the block? Solution

  13. Example: A particle of mass m = 5.00 kg is released from point A and slides on the frictionless track shown in Figure. Determine (a) the particle’s speed at points B and C and (b) the net work done by the gravitational force in moving the particle from A to C.

  14. (a) the particle’s speed at points B Find the particle’s speed at points C then find the work from the relation

  15. Work Done by Nonconservative Forces Nonconservative forces change the amount of mechanical energy in a system. Wnc = work done by nonconservative force

  16. Problem: Children and sled with mass of 50 kg slide down a hill with a height of 0.46 m.  If the sled starts from rest and has a speed of 2.6 m/s at the bottom, how much thermal energy is lost due to friction (i.e. what is the work that friction does)?  If the hill has an angle of 20° above the horizontal what was the frictional force. Since vi = 0, and hf = 0,

  17. The force done by friction is determined from;

  18. Example: A child of mass m rides on an irregularly curved slide of height h = 2.00 m, as shown in Figure. The child starts from rest at the top. (A) Determine his speed at the bottom, assuming no friction is present.

  19. (B) If a force of kinetic friction acts on the child, how much mechanical energy does the system lose? Assume that vf =3.00 m/s and m = 20.0 kg.

  20. Example: A block having a mass of 0.80kg is given an initial velocity vA = 1.2m/s to the right and collides with a spring of negligible mass and force constant k = 50N/m, as shown in Figure. (A) Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision.

  21. (B) Suppose a constant force of kinetic friction acts between the block and the surface, with = 0.50. If the speed of the block at the moment it collides with the spring is vA = 1.2 m/s, what is the maximum compression xC in the spring?

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