work-kinetic energy ( ) = ( ) - lsu

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

Special Case: Work done by Gravitational Force

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Wg ≡ F g • d x ∫ =

F g • d

where Fg = mg( ) − ˆ j ( ) and g = 9.81m /s 2

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.

What is the change in KE due to Gravitational Force ?

If the only force acting on an object is Gravitational Force then,

If an object is displaced upward (Δ y positive), the change in Kinetic Energy is negative (it slows down).

If an object is displaced downward (Δy negative), the change in Kinetic Energy is positive (it speeds up).

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Wnet = Wg = F g • d ( ) = ΔKE = KE f − KEi( )

Careful with the Notations…

WORK done by Gravitational Force

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Wg ≡ F g • d x ∫ =

F g • d

where Fg = mg( ) − ˆ j ( ) and g = 9.81m /s 2

W = weight W = work

this could be a problem - just keep your wits about you

What work is needed to lift or lower an object?

In order to “lift” an object, we must apply an external force to counteract the gravitational force.

€

Wnet ≡Wg +Wext = ΔKE

If (i.e. ), then

€

ΔKE = 0

€

vf = vi

€

Wg = −Wext

If an object is displaced upward (Δy positive), then the work done by the External force (non-gravitational force) on the object is positive.

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

Problem 7-17 A helicopter lifts an astronaut of mass m vertically upward a distance d

from the ocean by means of a cable. The acceleration of the astronaut is g/8.

How much work is done on the astronaut by: a) the force from the helicopter and b) the gravitational force on him?

What are c) the kinetic energy and d) the speed of the astronaut just before he reaches the helicopter.

Example You are moving out of town and need to load a heavy box of mass m into the back of the moving truck.

You can either pick the box straight up and set it in the truck (a height h above the ground) or slide the box a distance L up a ramp into the back of the truck. Assume the ramp has good rollers on it, so that it is frictionless.

Which way requires more work?

Which way requires a greater force? L

h

2 Guys and a Truck

(800) WE LIFT 4U

θ

Work due to Friction�

WORK due to friction is ALWAYS NEGATIVE - Energy is transferred OUT - Kinetic energy decreases or ΔKE < 0 (slow down)

Where did the energy go? THERMAL/Sound

Special Case: Work done by a Spring Force

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F = force from the springk = spring constant (stiffness) - units N/m[ ] d = displacement from equilibrium ( x = 0 )

Note: the force is always directed to “restore” the equilibrium position

Work Done by Spring

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Wspring = F spring x( ) • d x ∫

= −kx( )dxx1

x2

∫ = −k xdxx1

x2

∫

= − 12 k( ) x 2[ ]x1

x2 = − 12 k x2

2 − x12( )

Note: Work done by spring is positive (negative) if block moves towards (away) equilibrium position. It is zero if the block ends up at the same distance from x=0

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F = −k

d Hooke’s Law

variable force

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




(b) 2 cm, 3 cm.

€

Ws =12kxi

2 −12kx fi

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

€

Wspring = 12 kxi

2 − 12 kx f

2

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:

€

v = dt

but:

P = F ⋅ vso:

€

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.

€

Wab,1 =Wab ,2

€

Wab = F (x)• d x

a

b

∫

€

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

€

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

€

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

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

€

ΔUspring = 12kx f

2 − 12kxi

2

€

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

work-kinetic energy ( ) = ( ) - lsu

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