lecturenotes-Feb10

# lecturenotes-Feb10 - Wab ,1 = Wab ,2 Wab ,1 = Wba ,2 Wab ,1...

This preview shows pages 1–11. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
P ave = W Δ t P = dW dt W ab ,1 = W ab ,2 W ab ,1 = W ba ,2 C-F W ab , 1 W ab , 2 Non-C-F W ab , 1 ≠− W ba , 2 P = F net v Power E mech = KE + U Δ U = W conservative F Δ E mech = 0 Potential Energy Mechanical Energy Conservation of Mechanical Energy
Potential energy Potential energy : Energy U which describes the confguration (or spatial arrangement) oF a system oF objects that exert conservative Forces on each other. It’s the stored energy in system. Gravitational Potential energy : [~associated with the state of separation] Δ U grav = mg ( ) dy = mg y f y i ( ) y i y f = mg Δ y If U grav ( y = 0) 0 then U grav ( y ) = mgy Elastic Potential energy : [~associated with the state of compression/tension of elastic object] Δ U spring = 1 2 kx f 2 1 2 kx i 2 If U spring ( x = 0) 0 then U spring ( x ) = 1 2 kx 2 Δ U = W Δ U grav if going up U grav if going down U spring if x goes or (any displacement) DeFnition

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Mechanical Energy Mechanical energy: Only conservative forces Isolated system : Assuming only internal forces (no external forces yet - Sec. 8.6) No external force from outside causes energy change inside E mech = KE + U W net = Δ KE Sec. 7 - 3 W conservative = −Δ U Sec. 8 -1 Δ E mech = 0 = Δ KE + U ( ) Δ KE = −Δ U
How to apply your knowledge to solve problems? When Friction is present, it always removes energy, so E(final) is less the E(initioal) E mec ( final ) = E mec ( initial ) Energy removed by work of friction K f + U f = K i + U i F f d Always think about the problem and the signs!! My preference is to always use Mechanical Energy E mec ( final ) = E mec ( initial ) K f + U f = K i + U i If there are no non-conservative forces involved,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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? Δ U grav = mg Δ y = 0 = −Δ KE Δ KE = 1 2 m v A 2 v 0 2 ( ) v A = v 0 Δ U grav = mg Δ y = mg ( h /2 h ) = −Δ KE Δ KE = 1 2 m v B 2 v 0 2 ( ) 1 2 v B 2 v 0 2 ( ) = g h /2 ( ) v B = gh + v 0 2 Δ U grav = mg Δ y = mg ( h max h ) = −Δ KE Δ KE = 1 2 m 0 v 0 2 ( ) mg ( h max h ) = 1 2 m 0 v 0 2 ( ) h max = v 0 2 2 g + h h max
Problem 8-4: Roller Coaster: Different way What is the speed of coaster at a) Point A b) Point B c) How high will it go on the last hill? E Mech = mv o 2 2 + mgh = mv a 2 2 + mgh v a = v 0 h max E mec =E kin +U E Mech = mv 0 2 2 + mgh E Mech = mv o 2 2 + mgh = mv b 2 2 + mg h 2 v b = v 0 2 + gh E Mech = mv o 2 2 + mgh = 0 + mgh max h max = v 0 2 2 g + h

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
slides below to give her the greatest possible speed when she reaches the bottom of the slide. Which one should she choose? 1. A
This is the end of the preview. Sign up to access the rest of the document.

## lecturenotes-Feb10 - Wab ,1 = Wab ,2 Wab ,1 = Wba ,2 Wab ,1...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online