Potensial Energy

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

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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

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