Work & kinetic energy

•  CAPA homework #5 due Tues at 10pm

•  MC Exam scores are available on D2L

•  Will not be available during reg. office hours 1-2pm today, but I am available between 2-4pm

•  Beginning material in Chapter 7 in H+R

Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/

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Midterm Multiple Choice

Quantized Results because only 14 questions

Long Answer Chain Problem

Dot (scalar) product One type of vector multiplication is the dot product or scalar product which produces a scalar from two vectors

Note that

so

Also,

where is the angle between the two vectors

Clicker question 1 Set frequency to BA

What is the dot product of the vector and the vector ?

€

F = (−2,0,0)

A.  -2 B.  3 C.  -4 D.  4 E. 

€

A ⋅ F = AxFx + AyFy + AzFz = −4 + 0 + 0 = −4

New concept: Work Applying a force causes acceleration and therefore over time will change the velocity Slightly different concept is applying a force over a displacement which is called work.

These horses are clearly doing work. They are exerting a force over a distance

Unit of work (and energy) is a joule (J) which is a newton·meter (N·m)

More precise definition of work

In measuring work, only the part of the force that is along the displacement vector counts as work.

Car Car

Imagine a stalled car with wheels pointed straight so it can only move from left to right. A force applied perpendicular to the way the car can move is useless and does no work

However, a force applied along the direction it can move will move the car. Work is done on the car.

Clicker question 2 Set frequency to BA

A 1000 kg car of mass is going around a level corner at 10 m/s. The corner has a radius of 100 m. The car travels 50 m while going around the corner. How much work is done on the car by the force of friction which is keeping the car going in a circle?

A.  0 J B.  500 J C.  5000 J D.  100000 J E.  None of the above The frictional force that keeps the car in a circle is always perpendicular to the velocity so it does no work

More on work Work is done by a force and on an object

Also, since the amount of work can be negative if the force and displacement are in opposite directions

Can also consider the total work on an object

It is the sum of the work of all the forces and It is also the work done by the net force

Example of work

10 kg box

A cat drags a 10 kg box at constant speed across a flat horizontal floor. The box/floor coefficient of kinetic friction is 0.6. If the cat drags the box 4 m, how much work is done by the forces? Since there is no acceleration in x or y

The normal force and weight are perpendicular to the displacement and contribute no work The work by the cat on the box is

The work by friction is the same but negative:

The total work is 0 as could be seen because the net force is 0

Clicker question 3 Set frequency to BA

Albert Einstein lowers a book of mass m downward a distance h at constant speed v. The work done by the force of gravity on the book is

A.  0 B.  mgh C.  –mgh D.  None of the above

Force and displacement are in the same direction (down) and so the work is positive and equal to force (mg) times the distance (h).

Clicker question 4 Set frequency to BA

Albert Einstein lowers a book of mass m downward a distance h at constant speed v. The work done by the force of Einstein’s hand on the book is

A.  0 B.  mgh C.  –mgh D.  None of the above

Force and displacement are in opposite directions and so the work is negative and equal to force (mg) times the distance (h).

Clicker question 5 Set frequency to BA

Albert Einstein lowers a book of mass m downward a distance h at constant speed v. The work done by the net force on the book is

A.  0 B.  mgh C.  –mgh D.  None of the above

No net force so total work done is 0. Work is done by the two individual forces but they cancel so no total work.

Clicker question 6 Set frequency to BA

Work of gravity is always –mgh. Signs matter!

You push a beer keg up a ramp with constant speed. Suppose you push parallel to the ramp, with force "F". The ramp travels a distance d along the ramp, ending at height h as shown.

How much work did GRAVITY do on the crate?

A) mg d B) mg d cosθ C) zero

D)  mg h (which is equal to mg d sinθ) E) None of these

Derivation of kinetic energy Consider a one-dimensional case where an object moves in the x direction a distance s. A constant net force F is applied in the same direction.

The object must have acceleration

From our 1D motion equations:

Multiplying by gives

From this we can figure out

Work-Energy Theorem We define kinetic energy as

Sometimes KE is used instead of just K

The work-energy theorem is just

This allows us to relate the work found from force times distance to a change in kinetic energy which can be used to determine the change in velocity

Example – pitcher throws a 145g baseball at a speed of

25m/s – how much work was done?

€

K =12mv 2 =

120.145g( ) 25m /s( )2 = 45J

Work by a varying force What if the force varies over the distance in which we are trying to calculate the work ? For now assume force and motion are in the same direction

For instance what if (true for a spring) Can divide up displacement into many small pieces with an approximately constant force at each point

x

F

s

ks

Then the work W=Fs is just the area under the curve which in this case is a triangle so

Area under the curve Can calculate things by determining the area under the curve

-2 -4

3 1

4 2

2 4

v (m/s)

t (s) v (m/s)

3

2 t (s)

which is the area under the curve (rectangle)

t

v

This also works for an arbitrary curve

Area is signed so in this example the total area is 0 (as expected since average velocity is 0).

so finding the area of the velocity vs time graph gives the displacement

Clicker question 8 Set frequency to BA

A force on an object is plotted versus the position of the object. What is the magnitude of the work done in moving from the origin to 5 meters?

A.  0 B.  5 J C.  10 J D.  25 J E.  50 J

position (m)

Force (N)

0 1 5

10

2 3 4

If you plot force versus distance the area under the curve is work . A right triangle has area of ½(base)(height) and here the base is 5 m and the height is 10 N.