Loading…Collecting status…Please wait…

YOU COME IN PRESENTATION OF THE 16 GROUP NOW.PLEASE USE YOUR WHOLE OF BELT CONCENTRATION AND

ATTANTION BE PATIENTLY !!!GET TO ENJOY IT...

The group 16The Member Names:Nur ‘ Aini (140210102006)Muh Ihsan Sholeh (140210102101)

POTENSIAL ENERGY AND CONSERVATION OF ENRGY

Work done on a system by an external force

Aplications of the hooke’s Law

Reading a potential energy curve

Potential Energy and Conservative

Energy

Conservative Energy

Reading a potential energy curve

Finding The Force Analytically

The Potential Energy Curve

F(x)

U(x)= -W = - F (x) x Turning Points

Neutral Equilibrium

Unstable Equilibrium

Equilibrium

U(x) + K(x) = Emec

U(x) = Emec

K(x) = 0Stable

Equilibrium

Potential Energy Curve

2.00 kg particle moves along an x axis in one-dimensional motion while a conservative force along that axis acts on it. The potential energy U(x) associated with a force is plotted. That is, if the particle were placed at any position between x = 0 and x = 7 m ,it would have the plotted value of U. At x 6,5 m the particle has velocity vo = -4 m/s i.

(a) Determine the particle’s speed at x1 = 4,5 m ! (b) Where is the particle’s turning point located ?(c) Evaluate the force acting on the particle when it is in the region 1,9 < x < 4,0 m !

Example Problem 1 :

The Potential Energy Curve

1

Calculations:

(a) At x = 6,5 m , the particle has kinetic energy

K0 = mvo2

= (2.00 kg)(4.00 m/s)2

= 16,0 J

Because the potential energy there is U = 0 , the mechanical energy is :

Emec = K0 + U0 = 16.0 J + 0 = 16.0 J

Example Problem 1 :

1 4 65 7

16

U (J)

x (m)

20

7 K0

K1

Emec = 16,0 J

This value Emec for is plotted as a horizontal line. From that figure we see thatat x = 4,5 m, the potential energy is U1 = 7,0 J. The kinetic energy K1 is the difference between Emec and U1 , SO :

K1 = Emec - U1

= 16.0 J - 7.0 J = 9.0 J

Because K1 = mv12, we find : v1 = 3 m/s

d

(b) the proportionality of distances

=

d = 2,08 m

x = 4,0 - 2,08 = 1,9 m

0

(c) we see that for the range 1,9 < x < 4,0 m , The force is :

F(x) = - dU(x)/dx

F(x) = -

F(x) = 4,3 N

1 4 65 7

16

U (J)

x (m)

20

7 K0

K1

Emec = 16,0 J

d

0

Work done on a system by an external force

No Friction Friction

W = ΔU + ΔKor,

W = ΔEmec

Newton’s Law II

F – fk = m.a

constant

v2 = vo2 + 2ad

Fd = ΔEmec + Fk d

Fd = ΔEmec + ΔEth

Example problem 2

A food shipper pushes a wood crate of cabbage heads (total mass m 14 kg) across a concrete floor with a constant horizontal force of magnitude 40 N. In a straight-line displacement of magnitude d 0.50 m, the speed of the crate decreases from v0 0.60 m/s to v 0.20 m/s.

(a) How much work is done by force , and on what system does it do the work ?(b) What is the increase Eth in the thermal energy of the crate and floor ?

Example problem 2

Calculation :

W = Fd cos θ= (40 N)(0.50 m) cos 0°= 20 J

(a) Because the applied force is constant, we can calculat the work it does by using W = Fd cos θ, so :

(b) We can relate Eth to the work W done by with the energy statement W = Emec + Eth for a system that involves friction.

Example problem 2

Eth = W = (mv2 - mv02)

W = m(v2 - v02)

= 20 J - (14 kg) [(0.20 m/s)2 - (0.60 m/s)2]

= 22.2 J

The change Emec in the crate’s mechanical energy is just the change in its kinetic energy because no potential energy changes occur, so :

Conservative Energy

Isolated System External force and ΔEint transfer

ΔEmec + ΔEth + ΔEint = 0

Emec1 = Emec2 ΔEmec = - ΔEth - ΔEint

Nonconservative force is working

Nonconservative force is not

working

ΔEmec = 0

External force

ΔK = Fd

ΔEint Transfer

ΔEbiochem

ΔU + ΔK = Fd

ΔU + ΔK = Fd cos θ

Power

Pavg = P =

Conservative of Energy

External Forces and Internal Energy Transfers

vo

d

F

v

θ

mv - mvo = Fxd

K - Ko = (F cos θ)d

ΔK = (F cos θ)d

ΔU + ΔK = (F cosθ)d

F

APLICATIONS OF THE HOOKE’S LAW

That’s All from us

THANK

YOU...