# Energy Conservation 1. Mechanical energy conservation For closed isolated system 2. Open system 3. Conservative and nonconservative forces Forces such.

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Energy Conservation 1. Mechanical energy conservation For closed isolated system 2. Open system 3. Conservative and nonconservative forces Forces such as gravity or the elastic force, for which the work dose not depend on the path taken but only on the initial and final position, are called conservative forces For conservative forces the work done on a closed path (a lop) is equal to zero Friction is a nonconservative force

Example: A block is moved from rest at point A to rest at point B. Which path requires the most work to be done on the object? A)The table is leveled and friction is present. Path 1 Path 2 Path 3All the same B) The table is tilted and frictionless. Path 1 Path 2 Path 3All the same 1 2 3 A B

Example: A hammer slides along 10 m down a 30  inclined roof and off into the yard, which is 7 m below the roof edge. Right before it hits the ground, its speed is 14.5 m/s. What is the coefficient of kinetic friction between the hammer and the roof? Δx = 10 m h = 7 m v = 14.5 m/s 30  This can be solved using Newton’s laws and kinematics, but it’s looooooooooooooooooooooooong. h’

Example: In the system below, a 200 g box is pushed 4 cm against a spring with k = 250 N/m and released. The box slides along a frictionless horizontal surface and then up an incline which makes an angle of 30  with respect to the horizontal. The coefficient of kinetic friction between the box and the incline is 0.2. How far along the incline is the box when its speed is half its maximum speed? d ?d ? Compression: x = 4 cm θ = 30  μ k = 0.2 m = 200 g k = 250 N/m v MAX v MAX /2 1) In the first part of the motion, mechanical energy is conserved.

2) For the whole process, mechanical energy E = K + U g + U elastic is not conserved due to friction: ΔE = W friction

1D) The force is minus the slope of the U (x) curve Relation between U and F (conservative force)

x U x x = 0, F = 0 dU/dx = 0 x U x x 0 dU/dx < 0 Spring

x U x x > 0, F < 0 dU/dx > 0 x U x x > 0, F < 0 and larger in magnitude dU/dx > 0 and steeper than before The force always points “downhill”!!!

A Example: Which of the force versus position graphs matches the potential energy function U(x)? U x F x x x F F B C + − Slope: 0 0 0 Force = − slope !

1D) Relation between U and F (conservative force) 2 and 3D) The force is minus the gradient of the function U (x,y,z) The force is minus the slope of the U (x) curve

Visualization of a gradient in 2D Think of a hilly terrain where U is the altitude. The negative gradient of U is a vector whose: Direction points down the hill in the direction water would flow from that location (i.e., in the steepest direction). Magnitude is the slope of the hill in that direction x y U x y UU UU

Example: Find the force exerted at point P (0,1,2) m if the potential energy associated with the force is:

Whenever F = 0 (ie, dU/dx = 0), we have equilibrium. x U xSxS xUxU xNxN x S, x U and x N are points of equilibrium Equilibrium

x U The force brings it back to the equilibrium point. stable The force pulls it away from the equilibrium point. unstable The force remains zero, so the particle stays at the new position, which is also an equilibrium position. neutral What happens if the particle moves some small dx away from the equilibrium point? Stable/unstable/neutral equilibrium Minimum = stable equilibrium Maximum = unstable equilibrium Force points “downhill” Turning points : E =U (so K = 0)

x U U MIN KE MAX x0x0 U MAX ( = E ) KE = 0, turn-around points Particle moves here Forbidden region (KE < 0) Forbidden region (KE < 0) –x0–x0 Energy Diagrams Example 1: A box attached to a spring on a horizontal, frictionless table is released at x = x 0 from rest.

Example 2: The box is brought to x = x 0 and pushed, so its initial velocity is v 0. xtxt –xt–xt New turn-around points. x U x0x0 (before) (now)

x U E xtxt –xt–xt U KE How much kinetic/potential energy does the system have at every point? U KE U = 0 KE = KE MAX = E U = U MAX = E KE = 0

U x A particle is subjected to the force associated with this potential. No other forces are exerted on the particle. Describe the motion of the particle in the following cases. Example: Potential with two pits.

1. The particle is released from rest at point A. U x A UAUA At M 1, U is minimum, so K (and speed) is maximum M1M1 The particle oscillates between A and B. B At x B, U = E, so K (and speed) is zero → turn around point Direction of force F E forbidden OK The particle is forbidden from x x B (K < 0)

2. The particle is released at point A with a small * initial velocity v 0. U x A UAUA E At M 1, U is minimum, so K (and speed) is maximum M1M1 The turn-around points are defined by K = 0, so U = E : points C and D. D C The particle oscillates between C and D. Direction of force F forbidden OK

The particle keeps moving in the +x direction (no oscillations). Direction of force F forbidden OK 3. The particle is released from rest at point G. U x G UGUG E At M 1, U is minimum, so K (and speed) is maximum M1M1

4. The particle is released from rest at point H. The particle has maximum speed at point: U x H A.M 1 B.M 2 C.M 3 M3M3 M1M1 M2M2 E J The particle oscillates between H and J. Direction of force F forbidden OK

5. The particle is released from rest at point K. U x K E From initial conditions, E = U K OK The particle oscillates between K and L. forbidden L K

6. The particle is released from rest at point M 1. U x M1M1 E From initial conditions, E = U M1 Equilibriu m Force = 0 with v = 0 → Direction of force F If someone pushes the particle slightly away from M 1, the force pushes it back. Stable equilibriu m

7. The particle is released from rest at point M 2. U x M2M2 E From initial conditions, E = U M2 Equilibrium Force = 0 with v = 0 → Direction of force F If someone pushes the particle slightly away from M 1, the force pushes it further away. Unstable equilibrium

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