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Lecture 14 Goals: Chapter 10 Understand the relationship between motion and energy Define Potential Energy in a Hooke’s Law spring Exploit conservation of energy principle in problem solving Understand spring potential energies Use energy diagrams Chapter 11 Understand the relationship between force, displacement and work Assignment: HW6 due Tuesday Oct. 26th For Monday: Read Ch. 11 1

-mg (yf – yi) = ½ m ( vyf2 -vyi2 ) Energy -mg Dy= ½ m (vy2 - vy02 ) -mg (yf – yi) = ½ m ( vyf2 -vyi2 ) A relationship between y- displacement and change in the y-speed squared Rearranging to give initial on the left and final on the right ½ m vyi2 + mgyi = ½ m vyf2 + mgyf We now define mgy = U as the “gravitational potential energy”

Energy ½ m vi2 + mgyi = ½ m vf2 + mgyf Notice that if we only consider gravity as the external force then the x and z velocities remain constant To ½ m vyi2 + mgyi = ½ m vyf2 + mgyf Add ½ m vxi2 + ½ m vzi2 and ½ m vxf2 + ½ m vzf2 ½ m vi2 + mgyi = ½ m vf2 + mgyf where vi2 ≡ vxi2 +vyi2 + vzi2 ½ m v2 = K terms are defined to be kinetic energies (A scalar quantity of motion)

What is the SI unit of energy A erg B dyne C N s D calorie E Joule Reading Quiz What is the SI unit of energy A erg B dyne C N s D calorie E Joule

Inelastic collision in 1-D: Example 1 A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. What is the initial energy of the system ? What is the final energy of the system ? Is energy conserved? x v V before after

Inelastic collision in 1-D: Example 1 What is the momentum of the bullet with speed v ? What is the initial energy of the system ? What is the final energy of the system ? Is momentum conserved (yes)? Is energy conserved? Examine Ebefore-Eafter v No! V x before after

Elastic vs. Inelastic Collisions A collision is said to be inelastic when “mechanical” energy ( K +U ) is not conserved before and after the collision. How, if no net Force then momentum will be conserved. Kbefore  Kafter E.g. car crashes on ice: Collisions where objects stick together A collision is said to be elastic when both energy & momentum are conserved before and after the collision. Kbefore = Kafter Carts colliding with a perfect spring, billiard balls, etc.

Inelastic collision: Where did that energy go?

Kinetic & Potential energies Kinetic energy, K = ½ mv2, is defined to be the large scale collective motion of one or a set of masses Potential energy, U, is defined to be the “hidden” energy in an object which, in principle, can be converted back to kinetic energy Mechanical energy, EMech, is defined to be the sum of U and K.

½ m vyi2 + mgyi = ½ m vyf2 + mgyf Energy If only “conservative” forces are present, the total mechanical energy (sum of potential, U, and kinetic energies, K) of a system is conserved For an object in a gravitational “field” ½ m vyi2 + mgyi = ½ m vyf2 + mgyf U ≡ mgy K ≡ ½ mv2 Emech = K + U Emech = K + U = constant K and U may change, but Emech = K + U remains a fixed value. Emech is called “mechanical energy”

Example of a conservative system: The simple pendulum. Suppose we release a mass m from rest a distance h1 above its lowest possible point. What is the maximum speed of the mass and where does this happen ? To what height h2 does it rise on the other side ? v h1 h2 m

Example: The simple pendulum. What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum. y y=h1 y=0

Example: The simple pendulum. What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum E = mgh1 at top E = mgh1 = ½ mv2 at bottom of the swing y y=h1 h1 y=0 v

Example: The simple pendulum. To what height h2 does it rise on the other side? E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point. E = mgh1 = mgh2 or h1 = h2 y y=h1=h2 y=0

Potential Energy, Energy Transfer and Path A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless The ball is dropped The ball slides down a straight incline The ball slides down a curved incline After traveling a vertical distance h, how do the three speeds compare? h 1 3 2 (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell

Potential Energy, Energy Transfer and Path A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless The ball is dropped The ball slides down a straight incline The ball slides down a curved incline After traveling a vertical distance h, how do the three speeds compare? h 1 3 2 (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell

Example The Loop-the-Loop … again To complete the loop the loop, how high do we have to let the release the car? Condition for completing the loop the loop: Circular motion at the top of the loop (ac = v2 / R) Exploit the fact that E = U + K = constant ! (frictionless) h ? R Car has mass m Recall that “g” is the source of the centripetal acceleration and N just goes to zero is the limiting case. Also recall the minimum speed at the top is Ub=mgh U=mg2R y=0 (A) 2R (B) 3R (C) 5/2 R (D) 23/2 R U=0

Example The Loop-the-Loop … again Use E = K + U = constant mgh + 0 = mg 2R + ½ mv2 mgh = mg 2R + ½ mgR = 5/2 mgR h = 5/2 R h ? R

Variable force devices: Hooke’s Law Springs Springs are everywhere, (probe microscopes, DNA, an effective interaction between atoms) In this spring, the magnitude of the force increases as the spring is further compressed (a displacement). Hooke’s Law, Fs = - k Du Ds is the amount the spring is stretched or compressed from it resting position. Rest or equilibrium position F Du

Exercise Hooke’s Law 8 m 9 m What is the spring constant “k” ? 50 kg (A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m

SF = 0 = Fs – mg = k Du - mg Exercise 2 Hooke’s Law Fspring mg 8 m 9 m What is the spring constant “k” ? SF = 0 = Fs – mg = k Du - mg Use k = mg/Du = 500 N / 1.0 m Fspring 50 kg (A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m mg

F vs. Dx relation for a foot arch: Force (N) Displacement (mm)

F-Dx relation for a single DNA molecule

Measurement technique: optical tweezers http://phet.colorado.edu/sims/optical-tweezers/stretching-dna.jnlp

Force vs. Energy for a Hooke’s Law spring F = - k (x – xequilibrium) F = ma = m dv/dt = m (dv/dx dx/dt) = m dv/dx v = mv dv/dx So - k (x – xequilibrium) dx = mv dv Let u = x – xeq. & du = dx  m

Energy for a Hooke’s Law spring m Associate ½ ku2 with the “potential energy” of the spring Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

Energy diagrams Ball falling Spring/Mass system Emech Emech y In general: Ball falling Spring/Mass system Emech Energy K y U Emech K Energy U u = x - xeq

Equilibrium U U Example Spring: Fx = 0 => dU / dx = 0 for x=xeq The spring is in equilibrium position In general: dU / dx = 0  for ANY function establishes equilibrium U U stable equilibrium unstable equilibrium

Energy (with spring & gravity) 1 3 2 h -x mass: m Given m, g, h & k, how much does the spring compress? Emech = constant (only conservative forces) At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0 Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0

Energy (with spring & gravity) 1 3 2 h -x mass: m Given m, g, h & k, how much does the spring compress? Emech = constant (only conservative forces) At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0 Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0 Given m, g, h & k, how much does the spring compress? Em1 = Em3 = mgh = -mgx + ½ kx2  Solve ½ kx2 – mgx - mgh = 0

Energy (with spring & gravity) 1 3 2 h -x mass: m Given m, g, h & k, how much does the spring compress? Emech = constant (only conservative forces) At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0 Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0 Given m, g, h & k, how much does the spring compress? Em1 = Em3 = mgh = -mgx + ½ kx2  Solve ½ kx2 – mgx - mgh = 0

Energy (with spring & gravity) 1 mass: m 2 h 3 -x When is the child’s speed greatest? (A) At y1 (top of jump) (B) Between y1 & y2 (C) At y2 (child first contacts spring) (D) Between y2 & y3 (E) At y3 (maximum spring compression)

Energy (with spring & gravity) 1 2 h mg 3 kx -x When is the child’s speed greatest? (D) Between y2 & y3 A: Calc. soln. Find v vs. spring displacement then maximize (i.e., take derivative and then set to zero) B: Physics: As long as Fgravity > Fspring then speed is increasing Find where Fgravity- Fspring= 0  -mg = kxVmax or xVmax = -mg / k So mgh = Ug23 + Us23 + K23 = mg (-mg/k) + ½ k(-mg/k)2 + ½ mv2  2gh = 2(-mg2/k) + mg2/k + v2  2gh + mg2/k = vmax2

Recap Assignment: HW6 due Tuesday Oct. 26th For Monday: Read Ch. 11