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Physics 207, Lecture 13, Oct. 15 Goals: • Chapter 10 Understand the relationship between motion and energy Define Potential Energy in a Hooke’s Law spring Develop and exploit conservation of energy principle in problem solving • Chapter 11 Understand the relationship between force, displacement and work Assignment: HW6 due Wednesday, Oct. 22 For Monday: Read all of Chapter 11 Physics 207: Lecture 13, Pg 1 Energy If only “conservative” forces are present, the total 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 K ≡ ½ mv2 U ≡ mgy 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” Physics 207: Lecture 13, Pg 2 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 ? m h1 h2 v Physics 207: Lecture 13, Pg 3 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 Physics 207: Lecture 13, Pg 4 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 Physics 207: Lecture 13, Pg 5 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 Physics 207: Lecture 13, Pg 6 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) Use fact that E = U + K = constant ! Ub=mgh Recall that “g” is the source of Car has mass m the centripetal acceleration U=mg2R and N just goes to zero is the limiting case. h? Also recall the minimum speed v R at the top is gR y=0 Physics 207: Lecture 13, Pg 7 Example The Loop-the-Loop … again Use E = K + U = constant v mgh + 0 = mg 2R + ½ mv2 gR mgh = mg 2R + ½ mgR = 5/2 mgR h = 5/2 R h? R Physics 207: Lecture 13, Pg 8 Example Skateboard What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest? Assume we can treat the skateboarder as a “point” Assume zero of gravitational U is at bottom of the hill m = 25 kg .. R=10 m 30° R=10 m y=0 Physics 207: Lecture 13, Pg 9 Example Skateboard What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest? Assume we can treat the skateboarder as “point” Assume zero of gravitational U is at bottom of the hill m = 25 kg .. Use E = K + U = constant R=10 m Ebefore = Eafter 0 + m g R = ½ mv2 + mgR (1-sin 30°) 30° mgR/2 = ½ mv2 R=10 m gR = v2 v= (gR)½ v = (10 x 10)½ = 10 m/s Physics 207: Lecture 13, Pg 10 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 1. Ball is dropped 2. Ball slides down a straight incline 3. Ball slides down a curved incline After traveling a vertical distance h, how do the three speeds compare? 1 2 3 h (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell Physics 207: Lecture 13, Pg 11 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 1. The ball is dropped 2. The ball slides down a straight incline 3. The ball slides down a curved incline After traveling a vertical distance h, how do the three speeds compare? 1 2 3 h (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell Physics 207: Lecture 13, Pg 12 Example Skateboard Now what is the normal force on the skate boarder? N m = 25 kg 60° .. mg R=10 m S Fr = mar = m v2 / R 30° = N – mg cos 60° R=10 m N = m v2 /R + mg cos 60° N = 25 100 / 10 + 25 10 (0.87) N = 250 + 220 =470 Newtons Physics 207: Lecture 13, Pg 13 Elastic vs. Inelastic Collisions A collision is said to be elastic when energy as well as momentum is conserved before and after the collision. Kbefore = Kafter Carts colliding with a perfect spring, billiard balls, etc. vi Physics 207: Lecture 13, Pg 14 Elastic vs. Inelastic Collisions A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. Kbefore Kafter Car crashes, collisions where objects stick together, etc. Physics 207: Lecture 13, Pg 15 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 Physics 207: Lecture 13, Pg 16 Inelastic collision in 1-D: Example 1 What is the momentum of the bullet with speed v ? mv What is the initial energy of the system ? 1 1 2 mv v mv 2 2 What is the final energy of the system ? 1 (m M )V 2 2 Is momentum conserved (yes)? mv M 0 (m M )V Is energy conserved? Examine Ebefore-Eafter 1 1 1 1 m 1 m 2 mv 2 [( m M )V]V mv 2 (mv) 2 2 2 mM v mv 2 1 2 ( mM ) v No! V x before after Physics 207: Lecture 13, Pg 17 Variable force devices: Hooke’s Law Springs Springs are everywhere, (probe microscopes, DNA, an effective interaction between atoms) Rest or equilibrium position F Ds In this spring, the magnitude of the force increases as the spring is further compressed (a displacement). Hooke’s Law, Fs = - k Ds Ds is the amount the spring is stretched or compressed from it resting position. Physics 207: Lecture 13, Pg 19 Home Exercise Hooke’s Law 8m 9m What is the spring constant “k” ? Fspring SF = 0 = Fs – mg = k Ds - mg 50 kg Use k = mg/Ds = 5 N / 0.01 m (A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m mg Physics 207: Lecture 13, Pg 21 F-s relation for a single DNA molecule Physics 207: Lecture 13, Pg 22 Measurement technique: optical tweezers http://phet.colorado.edu/sims/optical-tweezers/stretching-dna.jnlp Physics 207: Lecture 13, Pg 23 Force vs. Energy for a Hooke’s Law spring F = - k (x – xequilibrium) F = ma = m dv/dt = m (dv/dx dx/dt) m = m dv/dx v = mv dv/dx xf vf So - k (x – xequilibrium) dx = mv dv ku du mv dv Let u = x – xeq. & du = dx xi vi 1 2 ku | 2 xf xi 1 2 2 vf mv vi | 1 2 kx kx 2 f 1 2 2 i 1 2 2 mv f 1 2 2 mvi 1 2 kx mv kx mv 2 i 1 2 2 i 1 2 2 f 1 2 2 f Physics 207: Lecture 13, Pg 24 Energy for a Hooke’s Law spring 1 2 kx mv kx mv 2 i 1 2 2 i 1 2 2 f 1 2 2 f Associate ½ kx2 with the “potential energy” of the spring m U si K i U sf K f Hooke’s Law springs are conservative so the mechanical energy is constant Physics 207: Lecture 13, Pg 25 Energy diagrams In general: Ball falling Spring/Mass system Emech Emech K K Energy Energy U U y s Physics 207: Lecture 13, Pg 26 Energy diagrams Spring/Mass/Gravity system spring alone Force y -mg Ug Emech K Energy spring & m K gravity Uspring UTotal y Physics 207: Lecture 13, Pg 27 Equilibrium Example Spring: Fx = 0 => dU / dx = 0 for x=0 The spring is in equilibrium position In general: dU / dx = 0 for ANY function establishes equilibrium U U stable equilibrium unstable equilibrium Physics 207: Lecture 13, Pg 28 Comment on Energy Conservation We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. Mechanical energy is lost: Heat (friction) Bending of metal and deformation Kinetic energy is not conserved by these non-conservative forces occurring during the collision ! Momentum along a specific direction is conserved when there are no external forces acting in this direction. In general, easier to satisfy conservation of momentum than energy conservation. Physics 207: Lecture 13, Pg 29 Physics 207, Lecture 13, Oct. 15 Assignment: HW6 due Wednesday For Monday: Read all of chapter 11 Physics 207: Lecture 13, Pg 30