Exam Scores Recent exam scores are shown as a histogram ordered simply by student ID number. Student ID number  What is the best approximation of the class average on this exam? A. 80 B. 60 C. 40 D. 20

avg = 85 Exam Scores The same scores are rearranged in ascending order below. What is the best approximation of the class average on this exam? A. 80B. 60 C. 40D. 20

velocity, v time (seconds) velocity, v time (seconds) Which of the graphs above could represent an object freely falling from rest? A.B. C.D.

velocity, v time (seconds) velocity, v time (seconds) For the moving object graphed at left A. v avg < v max v max 1212 B. v avg = v max 1212 C. v avg > v max 1212 v max 1212 v min For the moving object graphed at left A. v avg < v max 1212 B. v avg = v max 1212 C. v avg > v max 1212 v max 1212

WeightSupport (floor)

Adding all these supporting forces together some pull left, some pull right, some pull forward, some pull back all tend to pull UP! (a tug-of-war balancing)

Styrofoam bridge weighted at center Pressure applied to rigid glass bar

xx xx xx

Natural length xx Springs may be compressed (shortening their “natural length”) until the “restoring force” it counters with in an attempt to regain its original shape exactly balances the applied force. W F

Springs may be elongated (beyond their “natural length”) until the “restoring force” exactly balances the force pulling it open. Note: F   x xx

Compound bows use systems of pulleys and cams to maintain a fairly constant resistance force over the full distance drawn. Otherwise, for a more conventional recurve or simple longbow Force to hold in place distance bow is drawn 

Force Displacement,  x F  xF  x Also means: F = k  x “spring constant” in units of N/m The restoring force grows so many Newtons for every meter stretched.

The weight of a 1250 kg car is evenly distributed over 4 coil springs with strength k = N/m. When carrying 5 passengers (each, on average 73.0 kg) how much lower does the car ride? The empty car already compresses each spring: ¼ th total weight, empty car = m = 9.6 cm With everyone on board: = m = 12.4 cm It rides about 2.8 cm (a little more than 1 inch) lower.

snapped garage door spring sprung bicycle seat spring cracked suspension coil spring permanetly deformed Slinky!

Force Displacement,  x elastic range failure plastic deformation proportionality limit

Force Displacement,  x Work done against an elastic force starting with force  0 ending with force = k  x W = F avg d =  x F max 2 = k(  x)

Fd The “full draw weight’ of this bow is 20 pounds (90 Newtons) at the 24 inches (0.60 meter) that this archer has drawn it. What is its effective spring constant? How much work was done in drawing the bow back? If all the energy goes into the 400 grain (26 gram) arrow’s kinetic energy, what speed will it leave the bow with?

Total Weight

A tennis ball rebounds straight up from the ground with speed 4.8 m/sec. How high will it climb? It will climb until its speed drops to zero! final speed = 0 g =  9.8 m/sec 2 and in that much time it will rise a distance or since time up = time down the distance it falls from rest in 0.49 sec:

A tennis ball rebounds straight up from the ground with speed 4.8 m/sec. How high will it climb? It will climb until all its kinetic energy has transformed into gravitational potential energy! = = meter

A skier starts from rest down a slope with a 33.33% grade (it drops a foot for every 3 horizontally). h 3h3h The icy surface is nearly frictionless. How fast is he traveling by the bottom of the 15-meter horizontal drop? W The long way: The accelerating force down along the slope is in the same proportion to the weight W as the horizontal drop h is to the hypotenuse he rides: F The slope is long So it takes him During which he builds to a final speed:

A skier starts from rest down a slope with a 33.33% grade (it drops a foot for every 3 horizontally). h 3h3h The icy surface is nearly frictionless. How fast is he traveling by the bottom of the 15-meter horizontal drop? W The short way: The gravitational potential energy he has at the top of the slope will convert entirely to kinetic energy by the time he reaches the bottom. F = m/sec

h 3h3h3h3h 2h2h ½h½h 1.5h h The bottom of the run faces a slope with twice the grade (steepness). Neglecting friction, the skier has just enough energy to coast how high? A B C D

We know rubber tires are easily deformed by the enormous weight of the car they support…but not permanently. They regain their round shape when removed. This “spring-like” resiliency explains the rebound of all sorts of balls.

Racquetball rebounding from concrete. Tennis ball rebounding from concrete.

Airtrack bumper carts:

Notice that if the spring bumbers reverberate (ring) this would have to represent some energy that did not get returned to forward motion! This represents a fractional loss in kinetic energy! A purely ELASTIC COLLISION is defined as one which conserve kinetic energy. Unlike the stored potential of the compressed bumpers, this is not a “potential” energy that can ever be recovered as kinetic energy.

Look how a racquetball still undulates after leaving the floor! These vibrations are a wasted form of energy!