Work-Kinetic Energy Theorem

change in the kinetic energy of an object

net work done on the particle ( ) = ( )

Note: Work is the dot product of F and d

WF ≡

F • dx∫

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KE = 12mv

2

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Wnet = ΔKE = KE f −KEi

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Wg ≡ F g • d

x ∫ = F g • d = mgΔy

where Fg = mg( ) − ˆ j ( ) and g = 9.81m / s2

If an object is displaced upward (Δ y positive), then the work done by the gravitational force on the object is negative.

If an object is displaced downward (Δy negative), then the work done by the gravitational force on the object is positive.

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Wspring = F spring x( )• d

x ∫ = −

12

k x22 − x1

2( )

For the situation (Figure), the initial and final positions, respectively, along the x axis for the block are given below. Is the work done by the spring force on the block positive, negative or zero?

1.  Positive 2.  Negative 3.  Zero

Checkpoint Checkpoint #2

(a) -3 cm, 2 cm.

€

Ws =12kxi

2 −12kx fi

2

For the situation (Figure), the initial and final positions, respectively, along the x axis for the block are given below. Is the work done by the spring force on the block positive, negative or zero?

1.  Positive 2.  Negative 3.  Zero

Checkpoint Checkpoint #2

(b) 2 cm, 3 cm.

€

Ws =12kxi

2 −12kx fi

2

For the situation (Figure), the initial and final positions, respectively, along the x axis for the block are given below. Is the work done by the spring force on the block positive, negative or zero?

1.  Positive 2.  Negative 3.  Zero

Checkpoint Checkpoint #2

(c) -2 cm, 2 cm.

€

Ws =12kxi

2 −12kx fi

2

Sample Problem 7-8

A block of mass m slides across a horizontal frictionless counter with speed v0. It runs into and compresses the spring with spring constant k. When the block is momentarily stopped by the spring, by what distance d is the spring compressed?

Work by Spring force:

Work-Kinetic Energy theorem:

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Wnet = ΔKE

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Wspring = 12 kxi

2 − 12 kx f

2

HW #12 Problem

A block of mass m is dropped onto a spring. The block becomes attached to the spring and compresses it by distance d before momentarily stopping.

While the spring is compressed, what work is done on the block by: a) the gravitational force on it b) the spring force? c) What is the speed of the block just before it hits the spring? d) From what height h was the box dropped? e) How high will it go back.

Ch. 7 Problems HW#7: A block of mass m is attached to one end of a spring with spring constant k, whose other end is fixed. The block is initially at rest at the position where the spring is unstretched (x=0) when a constant horizontal force F in the positive x direction is applied. A plot of the resulting kinetic energy of the block versus its position x is shown. Find the equation relating the F to Ks.

Problem HW#8: A block of mass m is attached to one end of a spring with spring constant k, whose other end is fixed. The block is initially at rest at the position where the spring is unstretched (x=0) when a constant horizontal force F in the positive x direction is applied. a)   Where will it stop? b)   What is the work done by the applied force? c)  What is the work done by the spring? d)   Where is the block when the Kinetic Energy is max. e)  Value of Max. Kinetic energy.

Power • Power = the rate at which work is done by a force. • Average power is work W done in time Δt

• The instantaneous rate of doing work (instantaneous power)

• Units: Watt [W] 1 W = 1J/s

1 horsepower = 1 hp = 746 W 1 kW-hour = 3.6 MJ

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Pave =WΔt

P =dWdt

• Power from the time-independent force and velocity:

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P =dWdt

=d F • x ( )dt

= F •

d x ( )dt

= F • v Instantaneous power !

Power • The rate at which work is done by a force is power. • Average power is work W done in time Δt

• Units: Watt [W] 1 W = 1J/s

1 horsepower = 1 hp = 746 W €

P = Pave =Worktime

•  Power from the force and velocity:

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v = dt

but:

P = F ⋅ vso:

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P =Wt

=F→

• d→

t

An initially stationary crate of cheese (mass m) is pulled via a cable a distance d up a frictionless ramp of angle θ where it stops.

(a)  How much work WN is done on the crate by the Normal during the lift?

(b)  How much work Wg is done on the crate by the gravitational force during the lift?

(c)  How much work WT is done on the crate by the Tension during the lift?

(d)  If the speed of the moving crate were increased, how would the above answers change? What about the power?

Example: Crate of Cheese

WN = F→

N • d→

= 0

Wg = F→

g• d→

= −(mgsinθ)d = −mgh

WT = F→

T • d→

= (mgsinθ)d = mgh

P = F→

net • v→

• Chapter 7: What happens to the KE of when work is done on it. [ KE: “energy of motion” W: energy transfer via force ]

Conservative vs non-conservative forces •  Can you get back what you put in? Win = -Wout •  What happens when you reverse time?

Chapter 8: �Potential Energy & Conservation of Energy

Properties of Conservative Forces

• Net work done by a conservative force on an object moving around every closed path is zero.

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Wab,1 =Wab ,2

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Wab = F (x)• d x

a

b

∫

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Wab,1 = −Wba ,2&

Conservative forces - gravitational force

- spring force

Non-conservative forces - kinetic frictional force

(noise, heat,…)

Properties of Non-conservative Forces

• Net work done by a non-conservative force on an object moving around every closed path is non-zero.

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Wab,1 ≠Wab,2

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Wab = F (x)• d x

a

b

∫

Potential energy Potential energy: Energy U which describes the configuration (or spatial arrangement) of a system of objects that exert conservative forces on each other. It’s the stored energy in system.

Gravitational Potential energy: [~associated with the state of separation]

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ΔUgrav = − −mg( )dy = mg yf − yi( )yi

y f

∫ = mgΔy

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If Ugrav (y = 0) ≡ 0 then Ugrav (y) = mgy

Elastic Potential energy: [~associated with the state of compression/tension of elastic object]

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ΔUspring = 12kx f

2 − 12kxi

2

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If Uspring (x = 0) ≡ 0 then Uspring (x) = 12 kx

2

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ΔU = −W

ΔUgrav ↑ if going up

ΔUgrav ↓ if going down

ΔUspring ↑ if x goes ↑ or ↓ (any displacement)

Definition

Don’t forget…

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Wg = F g • d

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Wnet = ΔKE = KE f − KEi

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Wspring = 12 kxi

2 − 12 kx f

2

Work done by force (general)

Work by Gravitational force:

Work by Spring force:

Work-Kinetic Energy theorem:

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W = F ( x ) • d x ∫ =

F • d

Potential energy (if conservative force):

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ΔUgrav = mgΔy

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If Ugrav (y = 0) ≡ 0 then Ugrav (y) = mgy

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ΔUspring = 12 kx f

2 − 12 kxi

2

€

If Uspring (x = 0) ≡ 0 then Uspring (x) = 12 kx

2

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W = −ΔU