Knight Ch 10 - energy notes

# Knight Ch 10 - energy notes - Energy Ch 10 Lets look at the...

This preview shows pages 1–4. Sign up to view the full content.

Lets look at the equations of motion! This shows that the quantity before free fall is the equal to the same quantity after the fall. y 2a v v y 2 iy 2 fy + = i 2 iy f 2 fy gy 2 v gy 2 v + = + ) y 2g(y v i f 2 iy = Energy – Ch 10 ____________ Kinetic Energy & Potential Energy mg dt dv m ma F y net = = = dy dv v dt dy dy dv dt dv y y y y = = mg dy dv mv y y = Lets look the Newton’s second law in a freefall situation. F net = ____ Using the chain rule Placing this into the above equation Rearranging gives mgdy dv mv y y = = f i fy iy y y v v y y mgdy dv mv ____________ ____________ Kinetic Energy & Potential Energy mgdy dv mv y y = = f i fy iy y y v v y y mgdy dv mv f i fy iy y y v v 2 y mgy mv 2 1 = i f 2 iy 2 fy mgy mgy mv 2 1 mv 2 1 + = i 2 iy f 2 fy mgy mv 2 1 mgy mv 2 1 + = + This is an equation that shows the ___________ of energy as an object falls. It shows that the initial energy of the system is the same as the final energy of the ____________ _______________ Kinetic Energy & Potential Energy i 2 iy f 2 fy mgy mv 2 1 mgy mv 2 1 + = + There quantities have specific names. Kinetic Energy Gravitational Potential Energy Kinetic Energy is the energy of ______________. Potential energy is the energy of ___________.

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

View Full Document
Kinetic Energy & Potential Energy i 2 iy f 2 fy mgy mv 2 1 mgy mv 2 1 + = + Note: This equation can now be written as The total __________ of the system is not changed by freefall! gi i gf f U K U K + = + The unit of energy is called the Joule 1 joule = 1 J = 1 kgm 2 /s 2 Zero of Potential Energy The zero of potential energy is __________. You may set it to be wherever you like. We will typically take zero to be at the ______ point in the motion so that all of the height are positive. This also allows the potential energy of the object to be zero at the lowest point. Skiing down a Frictionless Hill A skier on ski’s weights 500N. They move down the slope a vertical distance of 100m. Find the speed of the skier at the bottom of the slope assuming his initial velocity is 10.0 m/s. 10.0 m/s 100m θ Skiing down a Frictionless Hill v i = 10.0 m/s v f = ? y i = 100m y f = 0m U i + K i = _____________ mgy i + ½ mv i 2 = mgy f + ½ mv f 2 Cancellation of the masses yields: ______ + ½ v i 2 = gy f + _______ (9.80m/s 2 )(100m) + ½ (10.0m/s) 2 = ½ v f 2 v f = [2[50m 2 /s 2 + 980m 2 /s 2 ]] 1/2 v f = 45 m/s Note: does not depend on ______ of hill!!!
Whee H R x Roller Coaster of Death!!!

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

Knight Ch 10 - energy notes - Energy Ch 10 Lets look at the...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online