Physics - Ch 5 Flashcards by Hannah Griffis

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

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Energy

The ability to change or cause change

No energy = no change

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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

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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

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H =

H = V² / 2g

H = (Vsinθ)² / 2g because height = vertical = y = sin

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V = (4)

V = √2gh (conservation of energy)

V = √2g (h-μkgcosθd)

V = d/t

V = mgh - Wfriction = 1/2mv²

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X and y components (2)

Dy = dsinθ 
Dx = dcosθ

Wx = mgcosθ
Wy = mgsinθ

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Equations with θ (2)

Mgsinθ = ma (force)

A = gsinθ

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μk = coefficient of friction

Ff = (force of friction)

(2)

μk = Ff/mg

Ff = μkFn –> μkmgcos

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Law of conservation of energy equations (2)

Mgh = 1/2mv²

Mgh - μkFfd = 1/2mv²

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Wnet =

ΔKE =

Wnet = ΔKE = Fnetd(cosθ) = μkmgd(cosθ)

ΔKE = KEf - KEi

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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)

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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)

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Total energy =

Total energy = (1/2mv²)+(mgh)

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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

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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

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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)

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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

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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

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)

1. Ep + Eki = Ekf 2. Mgh + 1/2mvi² = 1/2mvf² 3. gh + 1/2Vi² = 1/2Vf² 4. 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 =

1. Mgh - Wf = 1/2mv² --> V = √2g (h- μcosθd) 2. Wf = Ffd = μkFnd = μkmgcosθd 3. D = Wf / μkmgcosθd

How to find work with a flat surface and friction (4 steps)

1. F - Ff = 0 2. F = Ff 3. F = μkmg 4. 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: 1. Ep=mgh 2. Mgh=1/2mv² 3. Subtract 100 from % of energy lost 4. Take percent in decimal form from step 3 and attach to Mgh side 5. (#)mgh=1/2mv² 6. Solve for V Part 2: 1. Ek = 1/2mv² 2. Mgh=1/2mv² 3. Subtract 100 from % of energy lost 4. Take percent in decimal from from step 3 and attach to 1/2mv² side 5. Mgh = (#)1/2mv² 6. 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)

1. W = fd --> w=mgsinθd 2. 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)

1. (F+2F) - (mgsinθ + μkmgcosθ) = ma | 2. Solve for F

How to find time with power? (2)

1. 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