A rocket test projectile is launched skyward at

A rocket test projectile is launched skyward at 88 m/sec (198 mi/hr). How high does it go? vo = 88 ? m/sec vfinal = ? a = -9. 8 ? m/sec 2 time to reach peak, t = ? height reached, x=? When does it peak? What has happened by that point? v=0 = 9 sec So = 792 m – 396. 9 m = 395 m (1/4 mile)

A ball is thrown straight upward and caught when it returns to the height from which it was released. 1. At its peak position, the ball’s A. instantaneous velocity is maximum. B. instantaneous velocity is zero. C. instantaneous acceleration is zero. D. both B & C are true. 2. The time to fall back from its peak position is A. greater than B. equal to C. less than the time it took to climb that high. 3. The speed it builds up to downward by the moment it is caught is A. greater than B. equal to C. less than the speed it was thrown upward with.

For objects for which air friction is negligible, time up = time down speed down = speed up

Two spheres of identical mass are Released when the mechanism above Is triggered. One sphere is launched Horizontally by a spring, the other is simultaneously dropped from rest. A. The launched sphere will reach the ground first. B. Both spheres touch ground at the same time. C. The dropped sphere reaches the ground first.

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Galileo’s critics argued that if the earth moved:

A stone is flung horizontally at 8 m/sec from a point 1100 meters above the base of a cliff. How far away from the cliff does it land? or +1100 m +9. 80 m/s Careful!!! Actually ay = -9. 8 m/s 2 ax = ? 0 Its y-component of motion is like dropping from rest! VERTICALLY t = 15 sec HORIZONTALLY x = 120 m

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Weight Support (floor)

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

Styrofoam bridge weighted at center Pressure applied to rigid glass bar

on If. What this forces trunk isact moved this storage trunk? across the floor at constant velocity, Push Friction Support of floor (distributed among the 4 casters it sits on) Weight 1. the man’s PUSH must exceed the trunk’s WEIGHT. 2. the man’s PUSH must exceed the total friction forces. 3. all the above forces must exactly balance.

Forces all balance for constant velocity.

Imagine the work involved in sliding this crate from the loading dock to the center of the machine shop floor. Compare that to the task of pushing against the far wall (TWICE AS FAR)… TWICE AS MUCH …or to the task of pushing WORK? two identical crates together to the center of the room (TWICE AS MUCH).

Consider the work involved in lifting this heavy box to the bottom storage shelf. 2 h compare this to h doing it twice (once for each box) doing it once with BOTH boxes together TWICE lifting a single box to the AS MUCH upper shelf (TWICE AS FAR) WORK? HALF What about: half as high? AS MUCH half the weight? WORK?

Work Force used in performing it. Work distance over which work is done. Lifting weights is definitely work even by this physics definition! What about lowering weights back down?

How much work is being done balancing this tray in place? Work was done picking this briefcase up from the floor. How much work is being done in holding it still?

How much work does the cable do in supporting the bowling ball? T How much work does a crane do in holding its load in place above ground?

The crane lifts its load up at constant speed. 1. It lifts with a force > the load’s weight. 2. It lifts with a force = the load’s weight. 3. It lifts with a force < the load’s weight. When holding this load steady in place how much energy must its motor consume?

B C A D Work is done on the box during which stage(s)? A. Lifting box up from floor. B. Holding box above floor. C. Carrying box forward across floor. D. Setting box down gently to floor. E. At all stages: A, B, C, D. F. At stages A and C. G. At stages A and D H. At none of the stages illustrated.

Normal Force Weight

Normal Force Weight The unbalanced (net) force is parallel to the inclined surface of the road! Exactly the direction we know a car left in neutral will start to coast!

When work is done by a completely unbalanced force it can set objects in motion…or increase their speed. You push and it simply coasts away beyond your reach.

F d F d When an object moves in the direction of an unbalanced push it accelerates over the entire distance d. W=F d produces the acceleration a = F/m accelerating through the distance d in a time t: d = ½ at 2

Work is done on a wagon, initially at rest, by pushing it with an unbalanced force F over a distance D. With what final speed does it coast away? Final speed? With what acceleration a does it build speed? But for how long does it accelerate? So finally

But there’s another way to think about this W=F d F = ma d 1 2 = 2 at So the amount of work done is also equal to: W = ma 12 at 2 W = 12 ma 2 t 2 Then, just recall that the acceleration builds up, over t, to a final velocity: 0, if starting from rest! W= 1 2 mv 2 We consider this an expression of the energy possessed by a mass m when moving with a velocity v

Moving objects carry energy to scatter pins to knock down walls push snow aside In fact, energy is defined as the ability to perform work!

Raised weights also possess energy …potential energy… the potential to fall and build kinetic energy. A piledriver uses a raised weight to drive piles into the ground.

Raising a mass against gravity is doing work, even if there is no resulting final velocity.

Work done in lifting a mass to a height h: W=F d average force lifted with mg W = mgh In falling through a distance h an mass builds speed to what final kinetic energy? And in that time reaches So: KE = mgh

Normal Force Weight The unbalanced (net) force is parallel to the inclined surface of the road! Exactly the direction we know a car left in neutral will start to coast!

h d N W F x Pushing this box up along the ramp is 1. more work 2. less work 3. exactly the same work as simply lifting it straight up h. W F = d ? 1. h 2. N 3. x

W F = d h h d N W F x Work in simply lifting the box h Force in lifting distance lifted W h Work pushing box along the ramp is F d

F d A man, his briefcase in hand as shown, walks down the corridor from elevator to office door. He does 1. positive work on the briefcase. 2. no work on the briefcase. 3. negative work on the briefcase. The force F does is not along (nor back against) the direction of motion. Instead F and d are perpendicular! The force of support neither raises nor lowers the briefcase (neither increasing or decreasing its potential energy). It Neither speeds up nor slows down the briefcase (neither increasing nor decreasing the briefcases kinetic energy).

W F = d h h d N F W assume frictionless rollers on ramp x The box may slip down from its position on the shelf by sliding back down the ramp or falling straight down to the floor. By the time it reaches the floor it will reach 1. a higher final speed by falling. 2. a higher final speed by sliding. 3. the same final speed by either route.

W F = d h h d N F Falling: Sliding: then W

The merry-go-round is easier to push if all the riders 1. hang on at the outer edges. 2. huddle together at the center. 3. It makes little difference where they ride since its total mass that matters.