Download Work Done By Forces Conservative vs. Nonconservative Forces

Concepts Work Done By Forces ●
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W = ΔE W = F → * Δx→ Examples: ○ Person (you) ○ Friction →
F represents the part of the that is Parallel to the displacement Δx→
→
→
○ If F and Δx point in the same direction (are ​
parallel​
), then the Work done by the Force is positive →
○ If F →
antiparallel​
), then the and Δx point in opposite directions (are ​
Work by done the force is negative →
○ If F →
perpendicular​
, the ​
Work done by the force is zero and Δx are ​
Conservative vs. Nonconservative Forces ●
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A Conservative Force is a Force where, if an Object moves from point A to point B, then back from point B to point A, the work done by the force is zero. In other words ​
if the starting and ending position are the same, then the Work done by a Conservative Force is zero​
. ○ Gravity, Magnetism, and Electric Fields are Conservative ■ “I put things up and put them down” If starting and ending heights are the same, then work done by gravity is 0. A Nonconservative Force is a Force, where if the starting and ending positions are the same, the work can still be negative. ○ Friction is an example of a Nonconservative Force ■ If a push a heavy box in a full circle, the starting position and the ending position are the same, but the whole time friction was resisting motion. This is why ​
Work done by Friction is Always Negative​
, because it always “works” in the opposite direction of motion Velocity Has Direction! ●
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Remember that velocity can be both positive and negative An object that moves 5m/s to the right and then moves 5m/s to the left had a change in velocity of 10m/s (Not 0m/s). If that object had a mass of 2 kg, then its change in momentum was 20 kgm/s (Not 0kgm/s). Spring Energy is Always Positive ●
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P Es = 12k(Δx)2 1 k is always positive and (Δx)2 is positive because it is something squared. 2
Table for Different Variables in Physics ●
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The difference between a scalar and a vector is that a vector has direction, but a scalar does not. Scalars can be positive, like charge You don’t need to remember anything in this table except for first three rows. ○ Maybe just the Last Column. There is a shortcut for it, but I don’t want it to be confusing and it is difficult to remember without working with vectors in math ○ Get Units from Equations ○ Equations are on Equation Sheet Name Equation(s) Units Direction? Time NA s No, Scalar Displacement NA m Yes, Vector Mass NA kg No, Scalar Velocity v=
Δx
Δt
m
s
Yes, Vector Acceleration a=
Δv
Δt
m/s
s
Force F = ma F g = mg F f = μF N (kg)(m/s2) = kgm/s2 Also N , Newtons Yes, Vector P Eg P Eg = mgh (kg)(m/s2)(m) = kgm2/s2 = J No, Scalar KE K E = 12mv2 (kg)(m/s)2 = kgm2/s2 = J No, Scalar P Es P Es = 12k(Δx)2 P Es = J = kgm2/s2 J because Energy No, Scalar k F s = kΔx P Es = 12k(Δx)2 N = kgm/s2 = k(m) k = kg/s2 No, Scalar W By
(Force) W A = (F A )(Δx) (kgm/s2)(m) = kgm2/s2 Also J , Joules No, Scalar W ON
W ON
= ΔE J = kgm2/s2 Because Energy No Scalar W ON
= ∑Ef − ∑Ei = m/s = m/s2 Yes, Vector W By
(Object) W By
=− W ON
Energy ∑Ei ± W = ∑Ef J = kgm2/s2 Because Energy No, Scalar J = kgm2/s2 No, Scalar (With Forces/Work) (Forces not F g or F s ) ∑Ei = ∑Ef (Without Forces/Work) Power ρ = F *v ρ = ΔE
Δt ( ρ is Not actual symbol, this is rho for density) N (m/s) = (kgm/s2)(m/s) No, Scalar = kgm2/s3 Also W , Watts Momentum p = mv (kg)(m/s) = kgm/s Yes, Vector ∑pi ± F Δt = ∑pf (With Impulse) ∑pi = ∑pf (Without Impulse) Impulse I mpulse = F Δt I mpulse = Δp (N)(s) = (kgm/s2)(s) = kgm/s Yes, Vector Impact The F that causes Impulse N = kgm/s2 (Because Force) Yes, Vector Types of Collisions ●
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Momentum is Conserved for All Energy only Conserved for Elastic Collisions 1D Elastic Collision shortcut only works for 1 Dimensional Elastic Collisions. As long as those two conditions are satisfied, it works. Name Description Properties 1D Elastic (Billiard Balls) No Deformation Energy is Conserved No Energy lost in Deformation + v = v + v ​
v1i
Shortcut 1f
2i
2f
Inelastic (Bullet Travels Through Wood) Deformation Object passes through other E is not Conserved Mechanical Energy is lost in Deformation (Turns to Heat) Object Don’t Stick Perfectly Inelastic Objects Stick Mechanical Energy is lost in Deformation Explosions Both Objects start at rest Energy not Conserved Energy gained from Chemical Reaction, or something else Final momentum is 0 2D Collisions More than one Dimension Use Trig to Componentize all Vectors 2D Elastic Collisions 2D Billiard Balls No Deformation More than one Dimension Energy is Conserved (But not in just x or just y, only overall) Momentum is Conserved Impulse and Impact ●
According to the first table, ○ I mpulse = Δp ○ I mpulse = F Δt ○ Impact is either the maximum or the average value of F ○ If the initial velocity of the two objects is kept constant ■ The Initial momentum is constant (masses are also constant and p = mv ) ■ The final momentum is constant, (it is zero, because both objects stop at the end of the car accident or after the egg falls, etc. The questions will make this assumption) ■ So the change in momentum ( Δp ) is constant, and F Δt = Δp . So F Δt is constant. ■ If the collision lasts more time, then Δt is larger and F , the impact, must be smaller. ■ If the collision lasts less time, then Δt is smaller and F , the impact, must be be larger. ■ Impact and duration of Collision (time) are inversely proportional. Decreasing time increases impulse and visa versa 90° Shortcut ●
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On the applet you may have noticed that many of the 2D collisions resulted in final momentums/velocities that were perpendicular There are several conditions for this to be true ○ Only one object starts moving ○ The masses of the two objects have to be the same ○ The Collision has to be two Dimensional (which means that there is a glancing collision, the heights of the two objects when they collide have to different) ○ The collision has to be elastic ○ I don’t have a proof of this, and I am interested in proving it at some point. Impulses for One Object ●
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In collisions where Momentum is Conserved (all of them, unless there is an impact force acting on them) However, the momentum of each of the two objects could change between initial to final, as long as they change by opposite amounts If an Object bounces off something, the impulse is large ○ This is because the object had a positive momentum before it hit a wall, then had a negative momentum after it hit the wall ■ The object had a positive velocity before it hit the wall, so it had a positive momentum. After hitting the wall, it is moving in the opposite direction, so it has a negative velocity, so it has a positive momentum. ○ So the object had to experience a larger change in momentum (impulse) after bouncing off a wall. If an object enters a perfectly inelastic collision with another stationary object, it still moves in the same direction as before the collision. It is just moving slower, because it had to transfer some its momentum to the object it stuck to. ○ So the impulse for each object in an inelastic collision (where they stick) is lower. Power and Time ●
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ρ = ΔE
Δt The greater the change in energy, the larger that ΔE is, and the greater that the power is. The power is also increased if energy changes more quickly Predicting Results of a 1D Elastic Collision ●
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If one of two objects is stationary, and a second object collides with it elastically, the two objects could either move in the same direction or in opposite directions. If the moving object has more mass, then it will have a lot of momentum. It will have enough to keep on going after the collision and push the second object in the same direction. ○ It’s like if you shot a cannonball at an icecream sunday. The cannonball has a lot of momentum. It isn’t going to bounce off of the ice cream. It is going to destroy the ice cream by pushing it off the table. Except it can’t deform the ice cream because it is an elastic collision, so it doesn’t destroy the ice cream haha. ○ If you threw a pebble at a brick building, it would bounce off, because the pebble doesn’t have that much momentum, because it doesn’t have much mass. If it hit the wall, which has a lot of mass, it will only move the wall a little bit (it actually won’t move at all, because friction will prevent it from moving). The tiny bullet will bounce off, because even pushing a little bit on the wall will reduce the momentum of the pebble by a lot, enough to make it negative. Areas and Slopes ●
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Slope = Δy
Δx Slope units = y units
x units Slope = variable with those units Area = bh Area units = (x units)(y units) Area = variable with those units Bungee True/False ●
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Finding vmax
○ The object on the bungee is speeding up when acceleration points in the same direction as its velocity. ○ While it is moving down, gravity is causing it to speed up, because Fg points down. ○ After “the slack runs out”, F​
kicks in, and it points up. s​
○ As a long as Fg>Fs, it will continue to speed up. ○ Once Fs>Fg, however, the Net Force points up, which means that acceleration points up. Velocity and acceleration would now in opposite directions, so the object would start to slow down So the ​
maximum speed​
would happen ​
when Fs=Fg​
, which is ​
close, but Not at the point where the slack runs out​
. Finding amax
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Newton’s Second Law says “F=ma”. So the maximum value of a is where F is the biggest. Before “the slack runs out”, F is constant, because the only force acting on the object is Fg=mg. After the bungee runs out of slack, however, Fs kicks in. As delta x increases, Fs increases (because Fs=k*delta x). Fs reaches it’s maximum point at the bottom, where the bungee is stretched the most. At the bottom, Fs is equal to more than two times Fg. So the Net Force, Fs­Fg, is greater than Fg. So the Force is larger at the bottom than when the slack runs out. As a result, the maximum acceleration is at the bottom.