Clicker QuestionRoom Frequency BA A mass m is sliding on a frictionless plane tilted at an angle  as shown. The mass is moving up the plane because it was pushed up the plane by a spring a short time before (note that the spring is no longer in contact). Which of the following free-body diagrams most accurately represents the forces on the mass m at the moment shown in the diagram (after it was given an uphill push)? 59% 32% 1% 8%

Announcements CAPA Set #8 due tonight at 10 pm CAPA Set #9 now available, due next Friday Next week in Section: Lab #3 – Momentum Exam #2 – scores and solutions posted on CULearn detailed information on “exam info” web link Average = 67.4 out of 100 Stan. Dev. = 17.6 A range (85-100) B range (70-85) C range (55-70) D range (40-55) F range (0-40)

Clicker QuestionRoom Frequency BA A mass m is pulled to the right along a frictionless surface by a force of magnitude F at an angle , as shown. What is the magnitude of the normal force N exerted on the mass by the table? 31% 3% 15% 3% 48% mg N Fcos  Fsin  F net,y =N+Fsin  -mg=0 N=mg-Fsin 

“Do not try to bend the spoon — that's impossible. Instead, only try to realize the truth: there is no spoon.” This physics material is hard. It is particularly difficult to overcome preconceived notions that are backed up by years of observations. No force is required in the direction of the velocity. Keep working on these concepts and we are here to help.

Assume an isolated system (no external work). Assume no dissipation from friction (i.e. thermal energy generation) Roller Coaster Problem Clicker QuestionRoom Frequency BA Which of the following is true: A)Speed at position A = Speed at position B B)Speed at position A > Speed at position B C)Speed at position A < Speed at position B AB

Roller Coaster Problem Clicker QuestionRoom Frequency BA If the car is just sitting at the top of this hill: A)The net force on the car is upward. B)The net force on the car is downward C)The net force on the car is zero mg N F net,y = N-mg = ma y = 0

Roller Coaster Problem Clicker QuestionRoom Frequency BA If the car moving rightward at the top of this hill: A)The net force on the car is upward. B)The net force on the car is downward C)The net force on the car is zero mg N F net,y = N-mg = ma y = mv 2 /r v Approximately circular arc

ΔE mechanical + ΔE thermal = W external E mechanical = KE + PE ΔE thermal = thermal energy generated = -W friction W frict = -F friction Δx = -μ k N Δx W external = external work done on the system General Statement of Conservation of Energy System Energy can move from KE to PE to E thermal W external Can change energy of the system

KE i + PE i + W frict + W external = KE f + PE f W external > 0 if external work is done on the system W external < 0 if external work is done by the system W external = 0 if no external work is done on or by the system W frict < 0 if there is friction W frict = 0 if there is no friction or General Statement of Conservation of Energy ΔE mechanical + ΔE thermal = W external

Under what condition will the final mechanical energy be greater than the initial mechanical energy? A) W external > |W friction | B)W external < |W friction | C) W external = |W friction | D) It will never be greater than the initial mechanical energy E)It will always be greater than the initial mechanical energy Clicker QuestionRoom Frequency BA KE i + PE i + W frict + W external = KE f + PE f

Under what condition will W external be negative? A)If the external work is done on the system. B)If the external work is done by the system. C) It will never be negative. D) It will always be negative. Clicker QuestionRoom Frequency BA KE i + PE i + W frict + W external = KE f + PE f If there is work done by the system, the system loses energy to the surroundings that it does work on.

System = everything in the box One has to be careful about the definition of the “system” Friction between hippo and rug converts mechanical energy to thermal energy. However, no external work. All energy remains within the system (isolated). System = everything in the box Friction between hippo and rug converts mechanical energy to thermal energy. Friction force of rug on the hippo is now an external force. Thus energy leaves the system.

A mass m slides down a rough ramp of height h. Its initial speed is zero. Its final speed at the bottom of the ramp is v. h m Which of the following expressions gives the speed squared (v 2 ) of the block when it reaches the bottom of the ramp? Clicker QuestionRoom Frequency BA KE i + PE i + W frict + W external = KE f + PE f 0 + mgh + W frict + 0 = ½ mv 2 + 0

Find the work done by friction… Another Look h A mass m slides down a rough ramp of height h. Its initial speed is zero. Its final speed at the bottom of the ramp is v. KE i + PE i + W frict + W external = KE f + PE f 0 + mgh + W frict + 0 = ½ mv W frict = ½ mv 2 - mgh<0  (Mechanical Energy) = -W fric = mgh- ½mv 2

Another Look A mass m slides down a rough ramp of height h. Its initial speed is zero. Its final speed at the bottom of the ramp is v. h Find the work done by friction using another method… W fric = Force fric x displacement W fric =  k N x displacement W fric =  k (mgcos  ) x (h/sin  )  W fric =  k mgh cot(  )=  ME = ½ mv 2 - mgh

The effect of some forces is expressed as a Potential Energy: e.g., gravity, elastic forces produced by springs (later) These forces are said to be Conservative. They have the property that the work done depends only on the starting and ending points, not the path taken while the force is being applied. The work done by some other forces depends on the path taken: e.g., friction. Under the action of these forces, mechanical energy is lost. The friction force is Non-Conservative. Its effect cannot be expressed as a Potential Energy.

Clicker QuestionRoom Frequency BA A blue car coasts in neutral (no stepping on the gas or brakes) down the blue path. A red car coasts in neutral (no stepping on the gas or brakes) down the red path. If only gravity is acting, then A)The red car ends up with a larger velocity B)The blue car ends up with a larger velocity C)The two velocities are equal at the end

Force = -kx Hooke’s “Law” k = “spring constant” units = N/m x = displacement of spring Large k = stiff spring Small k = floppy spring Springs Spring in “relaxed” position exerts no force Compressed spring pushes back Force Stretched spring pulls back Force