Physics 1D03 - Lecture 19 Test TODAY at 9:30 am in CNH-104 !!!

Physics 1D03 - Lecture 19

Test TODAY at 9:30 am in CNH-104 !!!


• Review of scalar product of vectors• Work by a constant force• Work by a varying force• Example: a spring

Work and Energy


Work and EnergyWork and Energy

amF

Energy approach: Net work = increase in kinetic energy

- acceleration at any instant is caused by forces

- equivalent to Newton’s dynamics- scalars, not vectors - compares energies “before and after”

Newton’s approach:


)ABcos(

)A toparallel of (component x )A of (magnitude

BBA

B

A

The scalar product or dot product of two vectors gives a scalar result:

vector • vector = scalar

Math Review


ScaIar product and Cartesian components:

kBjBiBBkAjAiAA

zyx

zyxˆˆˆˆˆˆ

zzyyxx BABABABA

Then,

(note the right-hand-side is a single scalar)

To prove this, expand using the laws of arithmetic (distributive, commutative), and notice that

BA

since, i, j, k are mutually perpendicular

since they are parallel unit vectors1ˆˆˆˆˆˆ

0ˆˆˆˆˆˆ

kkjjii

kikjji

and


Work

Work by a constant force F during a displacement s:

d

IIF

F

F

Units : N • m = joule (J)

Work = (component of F parallel to motion) x (distance)

We can be write this as:

This is the “scalar product”, or “dot product”. Work is a scalar.If work is done on a system, W is positive (eg: lifting an object)If work is done by a system, W is negative (eg: object falling)

Work = F • d = Fdcos(θ)


Quiz

A constant force is applied to an object while it undergoes a displacement The work done by is :

N )ˆ3ˆ2ˆ1( kjiF

m. )ˆ2ˆ2ˆ2( kjid

F

N 12- e)

N 12 d)

J 12 c)

J 12- b)

J )ˆ6ˆ4ˆ2( a) kji


Example

A person walks down the stairs. Is the work done by gravity positive or negative ?

What if the person were to walk up the stairs ?


Example

You are pulling a sled with a force of 50N at an angle of 30o to the horizontal, for a distance of 25m. Determine the work done by you in pulling the sled.


Quiz

The two forces, P and Fg are constant as the block moves up the ramp. The total work done by these two forces combined is:

3 m

2 m

a) 20 J

b)

c) 40 J

J 32 J 1030 22

Fg = 5 N

P = 10 N


Example

2.5 m

fk = 50 N

w = 100 N

n

Fp = 120

A block is dragged 2.5 m up along the slope. Which forces do positive work?

A) negative work?

B) zero work?


Example

2.5 m

fk = 50 N

w = 100 N

n

Fp = 120

The block is dragged 2.5 m up along the slope of 37o. Find :

a) work done by Fp

b) work done by fk

c) work done by gravityd) work done by normal forcee) Total work on the block


a) Wp = (120 N)(2.5 m) = 300 J

b) Wf = (- 50 N)(2.5 m) = -125 J

c) Wg =

d) Wn = 0 ( motion)

J 150 )(37sin m) 2.5(N) 100( o

Total : 300 + (- 125) + (- 150) = 25 J

F=mg

d=2.5m sin(37o)

2.5m


5 min rest


Example: How much work is done to stretch a spring scale from zero to the 20-N mark (a distance of 10 cm)?We can’t just multiply “force times distance” because the force changes during the motion. Our definition of “work” is not complete.

Forces which are not constant:

ixfx

Varying force: split displacement into short segments over which F is nearly constant.

F(x)

x

FFor each small displacement x, the work done is approximately F(x) x, which is the area of the rectangle.

x


Work is the area (A) under a graph of force vs. distance

dxFWx

x

f

i

ixfx

Split displacement into short steps x over which F is nearly constant...

F(x)

xix

fx

F(x)

x

Take the limit as x 0 and the number of steps

xFW

We get the total work by adding up the work done in all the small steps. As we let x become small, this becomes the area under the curve, and the sum becomes an integral.

A


Determine the work done by a force as the particle moves from x=0 to x=6m:

x(m)

F(N)

0 1 2 3 4 5 6

5

Variable Force:


In 1D (motion along the x-axis): dxFWx

x

f

i

Another way to look at it: Suppose W(x) is the total work done in moving a particle to position x. The extra work to move it an additional small distance x is, approximately, W F(x) x.

Rearrange to getx

WxF

)(

In the limit as x goes to zero, dx

dWxF )(


An Ideal Spring

Hooke’s Law: The tension in a spring is proportional to the distance stretched.

or, F = kx

The spring constant k has units of N/m

Directions: The force exerted by the spring when it is stretched in the +x direction is opposite the direction of the stretch (it is a restoring force):

F = -kx and E=½kx2


Example:

Show that for a spring, E= ½ kx2.


A spring is hanging vertically. A student attaches a 0.100-kg mass to the end, and releases it from rest. The mass falls 50 cm, stretching the spring, before stopping and bouncing back.During the 50-cm descent, the total work done on the mass was:

a) zerob) 0.49 Jc) -0.49 Jd) none of the above

Quiz


Quiz

A physicist uses a spring cannon to shoot a ball at a gorilla. The cannon is loaded by compressing the spring 20 cm. The first 10 cm of compression requires work W. The work required for the next 10 cm (to increase the compression from 10 cm to 20 cm) would be:

a) Wb) 2Wc) 3Wd) 4W

Physics 1D03 - Lecture 19 Test TODAY at 9:30 am in CNH-104 !!!

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