Physics - Ch 5 Flashcards
Potential energy (8)
Energy of what might happen
Has potential to cause change
Joule (j)
Increases when mass, Gravity, or height increases
If height doubles Ep doubles
Ep = mgh
Linear
Depends on properties of the object and the objects interaction with the environment
Energy
The ability to change or cause change
No energy = no change
Kinetic energy (8)
Energy of motion
Joule (j)
Ek increases when mass increases
If mass doubles Ek doubles
If velocity doubles Ek quadruples
If velocity is tripled then it is gx the normal Ek
Ek = 1/2mv²
Ek = Ep - μfriction
Depends on mass and velocity
Law of Conservation of energy (3)
Energy can never be created or destroyed
Energy can be converted from form into another
Total energy before = total energy after
H =
H = V² / 2g
H = (Vsinθ)² / 2g because height = vertical = y = sin
V = (4)
V = √2gh (conservation of energy)
V = √2g (h-μkgcosθd)
V = d/t
V = mgh - Wfriction = 1/2mv²
X and y components (2)
Dy = dsinθ Dx = dcosθ
Wx = mgcosθ Wy = mgsinθ
Equations with θ (2)
Mgsinθ = ma (force)
A = gsinθ
μk = coefficient of friction
Ff = (force of friction)
(2)
μk = Ff/mg
Ff = μkFn –> μkmgcos
Law of conservation of energy equations (2)
Mgh = 1/2mv²
Mgh - μkFfd = 1/2mv²
Wnet =
ΔKE =
Wnet = ΔKE = Fnetd(cosθ) = μkmgd(cosθ)
ΔKE = KEf - KEi
Work = (10)
Work is the transfer of energy
Wp = fd (work done by person)
Wf = μkmgcosθd (work due to friction)
Wg = mgh (Work due to gravity)
Work is not done unless acted upon by an unbalanced force
Work is only done when a component of a force is parallel to a displacement
Scalar quantity – can be positive or negative
- positive if the force component is in same direction as displacement
- negative if the force component is in the direction opposite the displacement
Wx = mgcosθd Wy = mgsinθd
W = Fdcosθ (with a constant force)
How to find work done by gravity and work done by friction and Wnet? (3)
Work done by gravity = mgh
Work done by friction = μkmgcosθd
Wnet = work done towards motion/upwards - work done against motion)
Total energy =
Total energy = (1/2mv²)+(mgh)
Power (5)
The rate at which energy is transferred / work is done
P = w/t
P = F (d/t)
P= FV
Watts (w) —> 1 J per second
If θ = 0 or 90 (2)
If θ = 0 then cosθ = 1 and w = fd —> definition of work given earlier
If θ = 90 then cos 90 = 0 and w = 0 —> no work done
Gravitational potential energy (3)
The potential energy associated with an object due to its position relative to earth or some other gravitational source
Vx = 0 (h=0), only Vy exists
Depends on height and free-fall acceleration (neither are properties of an object)
Is there an x component to gravitational potential energy? If not why not? (2)
No because gravitational potential energy only operates in the y direction
Gravity doesn’t work in the x direction
When is mass irrelevant?
When potential energy = kinetic energy
How to find final KE ?
KE = PE - Ffriction
How to find Vf using the conservation of energy (4)
- PE + KE = KE bottom/end
- Mgh + 1/2mvi² = 1/2mvf²
- gh+1/2Vi² = 1/2mVf²
- Vf = √2(gh+1/2Vi²)
If someone is about to do a half pipe and they are 5m high (initially), how high will they get on the other side of the half pipe? Assuming no energy is lost to friction
5m
PE₁ =
PE₁ = PE₂ + Efriction
Can static friction do work? Explain.
No because static friction involves surfaced that are NOT moving
Also static friction does not have motion
How can a force do a negative work on an object?
If it opposes the motion of the object, it is a negative force and therefore does negative work
When is g positive/negative?
2
Positive - falling objects
Negative - raised objects
What are the unforeseen forces? Explain. (2)
Gravity / free-fall acceleration / friction = a force
Always an unseen force because of constant collisions in the air of air molecules
Mgh₁
Mgh₁ = mgh₂ + Efriction
Ef = friction of energy
Steps to find the final velocity using the law of conservation of energy (4)
- Ep + Eki = Ekf
- Mgh + 1/2mvi² = 1/2mvf²
- gh + 1/2Vi² = 1/2Vf²
- Vf = √2(gh + 1/2Vi²)
Wf = (3)
Friction due to work
Wf = Ffd (force of friction x displacement) –> μkmgcosθd
Ff = μkFn = mgcosθ
Similarities and differences between work and force? (2)
Similarity - both involve forces that cause accelerations
Difference - you can have a force that doesn’t cause motion but you can’t have work w/o motion
Explain how the law of conservation of energy explains the transfer of energy during an objects free-fall (2)
As the object falls it transfers some Ep into Ek –> h =
What is power and how is it related to work and forces ?
Power = rate at which work is being done and work involves forces
If Wnet = 0, is it moving or not?
Not enough info, it could be moving at a constant velocity or it could be standing still
Why do you need to find the x and y components to determine the work done? Example? (2)
If the force isn’t parallel to the objects motion; w=fd - the x and y components are used to find f
Ex. pulling a sled with a rope over my shoulder (pulling in x and y direction)
When would you set the force and opposing forces equal to each other?
If velocity is constant (a=o) and if there is no Fnet then set both sides equal to each other
Equations for energy and the conservation of energy (4)
? = ?
V =
H =
Mgh₁ =
Mgh = 1/2mv²
V (or Vf) = √2gh
H = V² / 2g
Mgh₁ = mgh₂ + Efriction
(Initial) (Final) (Friction in j)
Work and energy equations (4)
V =
Wf =
D =
- Mgh - Wf = 1/2mv² –> V = √2g (h- μcosθd)
- Wf = Ffd = μkFnd = μkmgcosθd
- D = Wf / μkmgcosθd
How to find work with a flat surface and friction (4 steps)
- F - Ff = 0
- F = Ff
- F = μkmg
- W = fd
How to find velocity when someone is going down a half pipe but loses a certain percent of energy due to friction? (6 steps)
And then how high does he go when he goes back up the half pipe if he also loses a percent of energy going up? (6 steps)
Part 1:
- Ep=mgh
- Mgh=1/2mv²
- Subtract 100 from % of energy lost
- Take percent in decimal form from step 3 and attach to Mgh side
- (#)mgh=1/2mv²
- Solve for V
Part 2:
- Ek = 1/2mv²
- Mgh=1/2mv²
- Subtract 100 from % of energy lost
- Take percent in decimal from from step 3 and attach to 1/2mv² side
- Mgh = (#)1/2mv²
- Solve for h₂
Use the concepts of work and energy to prove how the potential energy of a box at the top of a ramp comes from the work taken to push it there (3)
- W = fd –> w=mgsinθd
- Ep = mgh
W and Ep should be equal
If two people are pushing an object up a hill and one of them is pushing with twice the force as the second person, how do you find F? (2)
- (F+2F) - (mgsinθ + μkmgcosθ) = ma
2. Solve for F
How to find time with power? (2)
- T = √2d/a
2. P = w/t
Fnet= (2)
Fnet = μkmg
Fnet = mgcosθ + mgsinθ
How do you know when there is no work?
When there is no change in potential or kinetic energy
