P = dw dt. P = F net. = W Δt. Conservative Force: P ave. Net work done by a conservative force on an object moving around every closed path is zero

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2 Power Forces Conservative Force: P ave = W Δt P = dw dt P = F net v Net work done by a conservative force on an object moving around every closed path is zero Non-conservative Force: Net work done by a non-conservative force on an object moving around every closed path is non-zero

3 Potential energy: Energy U which describes the configuration (or spatial arrangement) of a system of objects that exert conservative forces on each other. It s the stored energy in system. ΔU = W Definition Gravitational Potential energy: [~associated with the state of separation] ΔU grav = y f y i ( mg)dy = mg( y f y ) i If U grav (y = 0) 0 = mgδy then U grav (y) = mgy Elastic Potential energy: [~associated with the state of compression/tension of elastic object] ΔU spring = 1 kx 2 2 f 1 kx 2 i2 If U spring (x = 0) 0 then U spring (x) = 1 2 kx 2 ΔU grav if going up ΔU grav if going down ΔU spring if x goes or (any displacement)

4 Work done by force (general) Work by Gravitational force: Don t forget W = F ( x ) dx W g = F g d = F d Work by Spring force: W spring = 1 kx 2 2 i 1 kx 2 2 f Work-Kinetic Energy theorem: W net = ΔKE = KE f KE i Potential energy (if conservative force): W = ΔU ΔU grav = mgδy If U grav (y = 0) 0 then U grav (y) = mgy ΔU spring = 1 kx 2 1 kx 2 f 2 i2 If U spring (x = 0) 0 then U spring (x) = 1 2 kx 2

5 Mechanical Energy Mechanical energy: E mech = KE + U Only conservative forces (gravity & spring) cause energy transfer (work) W net = ΔKE Sec. 7-3 W conservative = ΔU Sec. 8-1 Isolated system: Assuming only internal forces (no external forces yet - Sec. 8.6) No external force from outside causes energy change inside ΔE mech = 0 = Δ KE + U ( ) ΔKE = ΔU

6 How to apply your knowledge to solve problems? Step I: Choose a system Step II: Check if forces are conservative Step III: If so, apply ΔE mech = 0 ΔKE = ΔU ( ) = ( KE 2 + U 2 ) KE 1 + U 1

7 How to use it? Example: Pendulum ΔE mech = 0 ΔKE = ΔU ( ) = ( KE 2 + U 2 ) KE 1 + U 1 E mech = KE + U = constant [( ) + ( mgy) ] at all times = constant = 1 2 mv2 Assumes no friction (non-conservative, external force)

8 Example of Conservation of Mechanical Energy Consider a 100 kg bobsled starting with v o = 0 on a frictionless track. Compute the KE, PE, and total E at the labeled points. PE = mgh KE = 1 2 mv2 E mech = KE + PE E mech =KE+PE KE PE 588,000 J 0 J 588,000 J Height h 600 m 588,000 J 196,000 J 392,000 J 400 m 588,000 J 392,000 J 196,000 J 200 m 588,000 J 588,000 J 0 J 0 m

9 Example of Conservation of Mechanical Energy Consider a 100 kg bobsled starting with v o = 0 on a frictionless track. Compute the KE, PE, and total E at the labeled points. PE = mgh KE = 1 2 mv2 E mech = KE + PE New Zero E mech =KE+PE KE PE 0 J 0 J O 0 J Height h 0 m 0 J 196,000 J -196,000 J -200 m 0 J 392,000 J -392,000 J -400 m 0 J 588,000 J -588,000 J -600 m

10 Question Question 8-2 Three identical projectiles are fired at different launch angles and with different initial velocities. Which projectile has the greatest potential energy when it s at the peak in its trajectory? 1. A 2. B 3. C 4. All are equal Can you tell me anything about the kinetic energies?

11 Problem 8-4: Roller Coaster What is the speed of coaster at a) Point A b) Point B c) How high will it go on the last hill?

12 Problem 8-12: Roller Coaster h max What is the speed of coaster at a) Point A ΔU grav = mgδy = 0 = ΔKE ΔKE = 1 2 m v 2 2 ( A v 0 ) v A = v 0 b) Point B ΔU grav = mgδy = mg(h /2 h) = ΔKE ( ) ΔKE = 1 m v B v v 2 ( B2 v 0 ) = g h /2 ( ) v B = gh + v 0 2 c) How high will it go on the last hill? ΔU grav = mgδy = mg(h max h) = ΔKE ( ) ( ) ΔKE = 1 m 0 v mg(h max h) = 1 m 0 v h max = v 2 0 2g + h

13 Question Question 8-4 A block initially at rest slides down a frictionless ramp and attains a speed of v at the bottom. To achieve a speed of 2v, how many times as high must a new ramp be? KE f = KE i + PE i PE f = 0 + mgh 1 mv 2 2 f = mgh v f = 2gh

14 Question Question 8-3 A young girl wishes to select one of the frictionless playground slides below to give her the greatest possible speed when she reaches the bottom of the slide. Which one should she choose? A B C D 1. A 2. B 3. C 4. D 5. Any of them

15 Demo ΔE mech = 0 ΔKE = ΔU ( ) = ( KE 2 + U 2 ) KE 1 + U 1

16 Example A block of mass kg is given an initial velocity V a = 1.20 m/sec to the right and collides with a light spring of force (spring constant k=50.0 N/m) as seen in the figure. If the surface is frictionless, calculate the maximium compression of the spring after the collision.

17 A small block of mass m can slide along the frictionless loop-the-loop. The block is released from rest at point P, at a height h = 5R above the bottom of the loop. a) How much work does the gravitational force do on the block as the block travels from point P to point Q? b) If U = 0 at the bottom, what is the potential energy when the block is at the top of the loop? c) If, instead of being released, the block is given some initial speed downward along the track, how do the above answers change? d) What is the minimum height h so that the block makes it around the loop?

18 Problem A small block of mass m can slide along the frictionless loopthe-loop. The block is released from rest at point P, at a height h = 5R above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point P to point Q? W g = F g d = mg( ˆ j ) j 5Rˆ j ) j ) ( ) ((Rˆ ) = ( mg( ˆ ) (( 4Rˆ j )) = 4mgR If U = 0 at the bottom, what is the potential energy when the block is at the top of the loop? W g = F g d = 0 = ΔU U top U bottom = W g = mg( ˆ j ) j 0Rˆ j ) [( ) ((2Rˆ )] = 2mgR If, instead of being released, the block is given some initial speed downward along the track, how do the above answers change? Do the answers depend on velocity? NO -> Answers don t change What is the minimum height h so that the block ΔU = ΔKE makes it around the loop? From forces (see prior lecture) y ˆ : F g = ( mg) = m v2 or v min = gr r ( mgh min ) mg(2r) ( ) = 1 m v 2 2 top 2 ( v initial ) ( ) + 2R h min = 1 2g v 2 top 0 h min = 1 2g gr ( ) + 2R = 5 2 R

19 HW #12 Problem A block of mass m is dropped onto a spring. The block becomes attached to the spring and compresses it by distance d before momentarily stopping. While the spring is compressed, what work is done on the block by: a) What is the speed of the block just before it hits the spring? b) From what height h was the box dropped? c) How high will it go back.

20 Potential Energy Curve Plot of U(x), the potential energy as a function of the a system with 1-D movement along x-axis: E mech = KE(x) + U(x) = constant KE(x) = E mech U(x) Equilibrium Points Turning point E mech = 3 J F(x) = du(x) dx E mech = 5 J E mech = 4 J E mech = 1 J Equilibrium positions: where slope of U(x) curve is zero [i.e. F(x) = 0 ; NO FORCE ] -> Neutral vs Unstable vs Stable Equilibrium KE=0 ; F=0 & if move left or right move back KE=0 ; F=0 but if move left or right force to move away KE=0 ; E mech =U stationary particle

21 Finding a conservative force from potential energy F (x) U(x) In 1-D: ΔU(x) = W cons = Fdx x 2 x 1 then where F is a slowly varying, internal force acting on a particle in system ΔU(x) F(x)Δx Now go backwards, say you know the change in potential energy at some point and you want to know the force at that point (in the differential limit) F(x) = du(x) dx F grav (y) = du grav (y) dy F spring (x) = du spring (x) dx ( = mgy ) dy = = mg 1 ( kx 2 2 ) = kx dx

22 Potential Energy Curve Plot of U(x), the potential energy as a function of the a system with 1-D movement along x-axis while a conservative force does work on it: F(x) = du(x) dx F(x) is negative slope of tangent to U(x)

23 Problem #40: The figure shows a potential energy curve U vs position. If U A =9.00J, U C =20.J, U D =24.0J. The particle is released the point on U at the point between 1 and 3 m with a kinetic energy of 4J and U B =12. J. What is the Kinetic energy and speed at x=3.5 m and x=6.5m? Where is the turning point on the left and right. 16.0J 12.0J E k =mv 2 /2 9.0 J