Physics 1ALecture 7B

"Energy and persistence conquer all things.”--Benjamin Franklin

Spring Potential EnergyFor every conservative force that can perform work there is a potential energy that can be placed in the conservation of mechanical energy equation.

Non-conservative forces will not have potential energies associated with them.

For example, if you are displacing a mass on a spring from equilibrium:

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ΔU = − F spring ⋅ d

r A

B

∫

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ΔU = − −kxˆ i ( ) ⋅ dxˆ i xi

x f

∫

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ΔU = kx( )dxxi

x f

∫ = k x2

2 xi

x f

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ΔU =k2x f2 − xi

2( ) =k2Δ x 2( )

Spring Potential Energy

Spring Potential EnergyExampleA block of mass 12.0kg slides from rest down a frictionless 35.0o incline and is stopped by a strong spring with k=3.00x104N/m. The block slides 3.00m from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?

AnswerFirst, you must define a coordinate system.Let’s choose the down the incline as the positive x-direction and let the lowest position be x = 0.

Spring Potential EnergyAre there any non-conservative forces at work in this problem?

No, the only forces performing work are gravitational force and spring force. So:

ΔEmec = ΔK + ΔU = 0

The block starts at rest and ends at rest such that ΔK = 0. This leaves us with:

ΔU = ΔUgrav + ΔUspring = 0

 mgΔh + (1/2)kΔ(x2) = 0

-mgΔh = (1/2)kΔ(x2)

Answer

Spring Potential EnergyTo calculate the change in height of the block use:Answer

Δh

35o

block

x=3.00mΔh = (3.00m) sin35o

Δh = –1.72m

Conceptual QuestionBlocks A and B, of equal mass, start from rest and slide down the two frictionless ramps shown below. Their speeds at the bottom are vA and vB. Which of the following equations regarding their velocities is true?

1m1m

30o 60o

A B

A) vA > vB

B) vA = vB

C) vA < vB

Mechanical EnergyIn a system where non-conservative forces can cause energy changes, Emec may not be constant for the system.

We change the principle of conservation of mechanical energy in the following way:

Wnc = ΔEmec = ΔK + ΔU

where Wnc is the work done by non-conservative forces.

When starting energy problems, the first question you need to ask is: Are non-conservative forces performing work on the system?

Work and EmecExampleA 10.0kg block is released from point A in the figure below. The track is frictionless except for the portion between points B and C, which has a length of 6.00m. The block travels down the track, hits a spring (with a spring constant of 2,250N/m), and compresses the spring 0.300m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction, μk, between the block and the rough surface.

Work and EmecChoose up as positive y with the lowest point to be y = 0.Are there any non-conservative forces at work in this problem?

Yes, the frictional force between points B and C. So:

ΔEmec = ΔK + ΔU = Wnc = Wfriction

The block starts at rest and ends at rest such that ΔK = 0. This leaves us with:

ΔU = ΔUgrav + ΔUspring = Wfriction

 mgΔh + (1/2)kΔ(x2) = –(Ffriction)dBC 

Answer

Work and EmecLooking at the left side of the equation:

mgΔh = (10kg)(9.8N/kg)(0m – 3m) = -294J

(1/2)k(Δx)2 = (1/2)(2,250N/m)(0.3m)2 = 101JGiving us:

-294J + 101J = -193J = –(Ffriction)dBC

Since the block is not accelerating in the y-direction while it is crossing the rough surface, we can say:

ΣFy = FN – Fg = 0

FN = Fg = mg

 Ffriction = μk FN = μk(mg) 

Answer

Work and Emec

This gives us:

193J = (Ffriction)dBC = (μk)mgdBC

Answer

Remember to think about if non-conservative forces are present before starting energy problems.

Either way you can try and apply the mechanical energy equation.

Conceptual QuestionA block slides down an inclined plane which makes an angle of 25o with respect to the horizontal. Friction is present. Which of the following forces will perform non-conservative work on the block, thereby reducing the block’s total mechanical energy?

h

A) Force of Gravity.

B) Normal Force.

C) Force of Friction.

D) All of the above.

E) Both B and C.

block

Δx

25o

Multiple ObjectsExampleThe masses shown to the right are connected by a massless string over a frictionless, massless pulley and are released from rest. Find (a) the velocity of the 7.0kg mass just before it hits the floor, (b) the maximum height above the ground reached by the 4.0kg mass.

AnswerFirst, you must define a coordinate system.Let’s choose the up as the positive y-direction and let the ground be y = 0.

Multiple ObjectsAre there any non-conservative forces at work in this problem?

No, the only force performing work on the system is the gravitational force. So:

ΔEmec = ΔK + ΔU = 0ΔK = –ΔU

The masses start at rest and have the same velocity just before the 7.0kg mass hits the ground such that:

ΔK = Kf,7kg + Kf,4kg

ΔK = (1/2)m7kg(vf)2 + (1/2)m4kg(vf)2

ΔK = (1/2)(m7kg + m4kg)(vf)2

Answer

Multiple ObjectsThe potential energy of the 7kg mass will decrease while the potential energy of the 4kg mass will increase.

ΔU = ΔU7kg + ΔU4kg

ΔU = –(m7kg)ghi,7kg + (m4kg)ghf,4kg

ΔU = –(7kg)(9.8N/kg)(5m) + (4kg)(9.8N/kg)(5m)ΔU = –343J + 196J = –147J

Going back to the conservation of energy equation:ΔK = –ΔU = +147J

ΔK = (1/2)(11kg)(vf)2 = 147J

Answer

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v f =2 147J( )11kg

= 5.2ms part (a)

Multiple ObjectsAfter the 7kg mass hits the ground, energy is no longer conserved as the ground imparts a non-conservative force on the system.

But, since the string will just go slack the 4kg mass will still have the same velocity upward which will now be affected by gravity.

Answer

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ΔKE4kg = −ΔPE4kg

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ΔKE4kg = KE f −KEi = 0 − 12m4kgvi

2

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ΔPE4kg = PE f − PEi = m4kgghf −m4kgghi

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ΔPE4kg = m4kgg hf − hi( )

Multiple Objects

Putting all of the numbers gives us:

Answer

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hf = 5.0m+5.2m/s( )2

2 9.8N/kg( )

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hf = 5.0m+1.4m = 6.4m

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− 12m4kgvi

2 = −m4kgg h f − hi( )

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vi2

2g= h f − hi

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hf = hi +vi2

2g

Going back to conservation of energy equation:

For Next Time (FNT)

Finish reading Chapter 7.

Start the homework for Chapter 7.