Work-Energy TheoremCompiled by

Mphiriseni Khwanda

The Work-Energy Theorem and Kinetic Energy: horizontally along the plane

Consider a constant net external force acting on an object.The object is displaced a distance s, in the same direction asthe net force.

The work is simply s

F

smasFW 2

212

2122

21

ofof mvmvvvmasmW

axvv of 222 2221

of vvax

When a net external force does work on and object, the kineticenergy of the object changes according to

2212

21

onet KEKEW omvmv

THE WORK-ENERGY THEOREM: Horizontally

Work-Energy Theorem : vertically

sFW cos

PE

hhmg

hhmgW o

0

gravity

hh foshence hh of

sbut

Hence the work done by the force of gravity is equal to minus the change in potential energy

Work-Energy Theorem: Combination of both Conservative and Nonconservative Forces

In normal situations both conservative and non-conservative forces act simultaneously on an object, so the work done by the net external force can be written as

nccnetWWW

KEKEKE of W net

PEPEPE fogravity foc mghmghWW

ncc WWW

ncW PEKE

THE WORK-ENERGY THEOREM becomes:

PEKE ncW ofof PEPEKEKEPEKE ncW

ooff PEKEPEKE ncW

of EE ncWIf the net work on an object by nonconservative forcesis zero, then its energy does not change:

of EE

4.13 The Conservation of Mechanical Energy

THE PRINCIPLE OF CONSERVATION OF MECHANICAL ENERGY

The total mechanical energy (E = KE + PE) of an objectremains constant as the object moves, provided that the net

work done by external nonconservative forces is

In simple terms, total mechanical energy before = total mechanical energy after

provided that the net work done by external nonconservative forces is zero

What is the catch?Condition for that to be true

zero.

6.5 The Conservation of Mechanical Energy

SUMMARY: WORK-ENERGY THEOREMCase 1: Horizontally

The work done by the net force (external, eg forces applied) on the object is equal to the change in the object’s kinetic energy.

(

Case 2: Vertically

The work done by the non-conservative forces (eg friction) on the object is equal the change in both the kinetic and potential energies of the object

The work done by the net force(external, eg force of gravity) on the object is equal to minus the change in the object’s potential energy 𝑊 𝐶=𝑭 𝒈 .𝒅=−∆𝑃𝐸=−mg(h−h0)

𝒎𝒈 .𝒅=−mg (h−h0)Case 3: The general case (the combination of the two)

𝑾 𝑵𝑪=∆𝑲𝑬+∆𝑷𝑬

𝑊 𝑁𝐶=(∑ 𝒇 ¿ .𝒅=𝝁𝒎𝒈 .𝒅 ¿❑=( 12 m𝑣❑2 −12m𝑣0

2)+(mgh−𝑚𝑔 h0)𝑊 𝑁𝐶=( 12 m𝑣❑

2 +mgh)−( 12m𝑣0

2+𝑚𝑔 h0) 𝑊 𝑁𝐶=𝐸−𝐸0

Hence , the net work done by non-conservative forces is equal to the change in the energy of the system.

If 𝐸=𝐸0

( 12 m𝑣❑2 +mgh)=( 1

2m𝑣0

2+𝑚𝑔h0)

Problem solving strategy

Problem statement

Non-conservative forces present?

YES NO

𝑊 𝑁𝐶=𝐸−𝐸0

𝑊 𝑁𝐶=( 12 m𝑣❑2 +mgh)−( 1

2m𝑣0

2+𝑚𝑔 h0)

𝐸=𝐸0

Substitute and calculate the unknowns

(∑ 𝒇 ¿ .𝒅¿❑=( 12m𝑣❑2 −12m𝑣0

2)+(mgh−𝑚𝑔h0)Where Ʃ f represents sum of all non-conservative forces present and d the distance covered during the application of non-conservative forces

Applications: Work-energy theorem

1. A ramp in an amusement park is frictionless. A

smooth object slides down the ramp and

comes down through a height h, What

distance d is necessary to stop the object on

the flat track if the coefficient of friction is μ.

Section PQ is frictionless (keyword)(Non-conservative forces absent)

Hence

𝐸𝑄=𝐸𝑃

Hence the total potential energy at p was converted to kinetic energy at Q

Section QR has friction

Non-conservative forces present

Hence

𝑊 𝑁𝐶=𝐸𝑅−𝐸𝑄

(∑ 𝒇 ¿ .𝒅¿❑=( 12m𝑣𝑅2 +mgh𝑅)−(1

2m𝑣𝑄

2 +𝑚𝑔h𝑄)𝝁𝒎𝒈 .𝑑=( 12 m(0)❑

2 +mgh𝑅)−( h𝑚𝑔 +𝑚𝑔h𝑄)

𝜇𝑚𝑔 .𝑑=mgh𝑅− h𝑚𝑔 −𝑚𝑔h𝑄 Note that =

𝜇𝑚𝑔 .𝑑=− h𝑚𝑔

𝒅=−𝒉𝝁

Roller Coaster activity

Supposed a roller coaster starts from rest at point A and moves without friction as shown in the diagram.

(a)How fast is it moving at points B, C and D?

(b)What constant acceleration must be applied at D to have it stop at E?

Activity

The diagram on shows A 0.41-kg block sliding from A to B along a frictionless surface. When the block reaches B, it continues to move along a horizontal surface BC where a kinetic frictional force acts. As a result, the block slows down, coming to rest at C. The kinetic energy of the block at A is 37J and heights above the ground.

(a)What is the kinetic energy of the block when it reaches B?

(b)How much work does the kinetic frictional force do during the BC segment trip?

Activity

A child of mass m is released from rest at the top of water slide, at a height h = 9 m above the bottom of the slide as shown on the right. Assuming that the slide is frictionless because of the water on it, find the child’s speed at the bottom of the slide.