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The law of conservation of mechanical energy PowerPoint Presentation

The law of conservation of mechanical energy

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The law of conservation of mechanical energy

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The law of conservation of mechanical energy

TOPIC: The law of conservation of mechanical energy

How is kinetic energy defined?

Kinetic energy is defined as Ek =1/2 mv2

How is gravitational energy is defined?

Gravitational potential energy is defined as Epg =mgh

How is elastic potential energy is defined?

Kinetic energy is defined as Eps=1/2kx2

1 、 The concept of mechanical energy

Mechanical energy is total of Ek and Ep.

We can express it as E=Ek+Ep

E=1/2mv2+mgh+1/2kx2

- 、The kinetic energy and potential energy converted into each other

For example:

(1) When an object is free fall

How does energy convert?

(2) When an object is thrown upward ,

what is the energy conversion?

(3) When the arrow is shot out by the bow,

what is the energy conversion?

P

1

2

Q

P Q

Path1：P→Q

A O B

Path2：P→B→Q

Conclusion: kinetic energy and potential energy can be converted into each other.

3、 The concept of conservative forces

An object moves from point P to point Q along path1 or path2,

The work done by gravity is the same.

WG = mgh

The work done by elastic force is the same.

Ws= 1/2kx22-1/2kx12

What conclusion can we come to according to the examples above?

A force is conservative force, if the work done by the force on an object moving between two points is independent of the path taken.

4、 The law of conservation of mechanical energy:

According to the theorem of kinetic energy ,

the work done by the resultant force equals the change of kinetic energy.

W=△Ek (1)

The resultant force is only gravity

W= mg (hi－hf ) (2)

The change of kinetic energy

△Ek= 1/2 mvf2－1/2mvi2 (3)

Hence mg (hi－hf )=1/2mvf2－1/2mvi2

Or 1/2mvi2+mghi=1/2mvf2+mghf

Illustration: We get this equation by free fall motion. Similarly, either an object is in rectilinear motion or in curvilinear motion, only the work was done by gravity. We can always get the same conclusion.

This equation can be given :

E=Eki+Epgi =Ekf +Epgf

This is the law of conservation of mechanical energy.

It states that the total mechanical energy of a system remains constant if only gravity does works.

Similarly, if only the elastic force does work, the law of

conservation of mechanical energy can be given

1/2mvi2 +1/2kxi2=1/2mvf2+1/2kxf2

or E=Eki +Epsi=Ekf+Epsf

Generally, if only conservative forces do work,

the law of conservation of mechanical energy states

that

the total mechanical energy of a system remains constant if the only force that does work is conservative force.

This is equivalent to the statement that

if the kinetic energy of a conservative system increases ( or decreases) by some amount , the potential energy must decrease ( or increase) the same amount.

It can be given E = Eki+∑Epi=Ekf + ∑Epf

Ekf - Eki = ∑Epi - ∑Epf

θ

5 、Application:

Example problem: A pendulum consists of a ball of mass 100g attached to a light cord of length 50cm. The ball is released near nose of a person from rest when the cord makes an angle 60° with the vertical ( ignoring air resistance)

A: Predict—Observe—Explain(POE)methods

(1)When the ball comes back, you predict what will happen?

( Can the ball hit the person’s nose?)

(2)Do this experiment and observe carefully the result of this experiment.

What is the result of this experiment?

(3) Explain the result of this experiment with the law of conservation of mechanical energy.

In the motion of the ball, two forces act on the ball. One is gravity and the other is the force of tension. Only gravity does work ( the direction of tension is perpendicular to the direction of the ball’s velocity, tension doesn’t do work), the mechanical energy is conserved. Its initial kinetic energy and its final kinetic energy are zero. The energy is entirely gravitational energy, so the ball comes back to original position a. As a result, the ball can’t hit the nose of the person.

B Calculation: What is the speed of the ball when it is at the lowest point b ?

Solution:

We choose to take the zero potential energy at point b.

Known

Unknown

M=100g=0.1kg

Eka=0

L=50cm=0.5m

Vb=?

Epb=0

θ = 600

Vb = 2ghab

= 2g L (1- cos600 )

= 2 10 0.5 (1 – 0.5 ) m/s

×

×

×

= 5 m/s

According to the law of conservation of mechanical energy

Eka+Epa=Ekb+Epb

hab =L –Lcosθ

0+mghab=1/2mvb2+0

6 Activity of students --------Ask and answer the

questions studied with each other .

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