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Physics: Work and Energy Concepts

The document discusses work done in lifting an object. It defines work as the force applied over the distance of displacement. For a constant force lifting an object, work equals the force times the height lifted. It also introduces the concept of gravitational potential energy and shows that the work done by the lifting force equals the change in both kinetic and potential energy.

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Leon Mathaios
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0% found this document useful (0 votes)
367 views15 pages

Physics: Work and Energy Concepts

The document discusses work done in lifting an object. It defines work as the force applied over the distance of displacement. For a constant force lifting an object, work equals the force times the height lifted. It also introduces the concept of gravitational potential energy and shows that the work done by the lifting force equals the change in both kinetic and potential energy.

Uploaded by

Leon Mathaios
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Work done in lifting an object

y=h A force F lifts the mass at constant speed


through a height h.
F
The displacement is h
a=0
The applied force in the direction of the
y=0 displacement is:

F = mg (no acceleration)
mg
The work done by the force F is:

W = Fh = mgh
But the kinetic energy has not changed the gravity force mg
has done an equal amount of negative work so that the net work
done on the mass is zero
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Work done in lifting an object

y = h Alternative view: define a different form of energy


Gravitational potential energy, PE = mgy
F

a=0 Define:

Mechanical energy = kinetic energy + potential energy


y=0
Mechanical energy, E = mv2/2 + mgy
mg
Then:

Work done by applied force, F, is (change in KE) + (change in PE)

So W = Fh = KE + PE

15
Check, using forces and acceleration
y=h v
Net upward force on the mass is F mg
F
Apply Newtons second law to find the acceleration:
a
F mg = ma,
y=0 vo
so, a = (F mg)/m

mg One of famous four equations

v2 = v20 + 2ah

So, v2 = v20 + 2(F mg)h/m


(m/2)
mv2/2 mv20/2 = Fh mgh That is, DKE = Fh DPE
Or, W = Fh = DKE + DPE
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Or, W = Fh = DKE + DPE

If there is no external force, W = 0 and

0 = DKE + DPE

so that DKE = DPE

As the mass falls and loses potential energy, it gains an equal


amount of kinetic energy.

Potential energy is converted into kinetic.

Energy is conserved overall.

17
6.26 m = 0.6 kg, yo = 6.1 m

Ball is caught at y = 1.5 m


mg

a) Work done on ball by its weight?

Weight force is in same direction as the displacement so,

Work = mg displacement = 0.6g (6.1 - 1.5 m) = 27 J

b) PE of ball relative to ground when released?

PE = mgyo = 0.6g (6.1 m) = 35.9 J

c) PE of ball when caught?

PE = mgy = 0.6g (1.5 m) = 8.8 J


18
d) How is the change in the balls PE related to the work done by
its weight?

Change in PE = mg(y - yo) (final minus initial)

Work done by weight = mg (displacement) = mg(yo - y) = PE

19
Conservation of Mechanical Energy

In the absence of applied forces and friction:

Work done by applied force = 0

So, 0 = (change in KE) + (change in PE)

And KE + PE = E = constant

20
Example:

No applied (i.e. external) forces

E = KE + PE = constant

KE = mv2/2
PE = mgy

So E = mv2/2 + mgy = constant, until the ball hits the ground


21
Check:

vx = v0 cos q = constant, in absence of air resistance

v2y = v20y 2g(y y0) = (v0 sin q)2 2g(y y0) when object is at height y

v2 = v2x + v2y = (v0 cos q)2 + (v0 sin q)2 2g(y y0)
v2 = v20 2g(y y0), as sin2 q + cos2 q = 1
(m/2)
So, mv20/2 + mgy0 = mv2/2 + mgy and KE + PE = constant

22
6.34
Find the maximum
height, H.

Ignore air resistance.

Conservation of mechanical energy: KE + PE = constant

At take-off, set y = 0: E = mv20/2 + 0

At highest point, y = H: E = mv2/2 + mgH

So, E = mv20/2 = mv2/2 + mgH


(v20 v2)/2 (142 132)/2
H= = = 1.38 m
g 9.8

23
6.38 v = 0 at highest point

y=
= yo

Find the speed of the particle at A (vo). There is no friction.

Conservation of mechanical energy: E = KE + PE = constant

At A: E = mv20/2 + mgy0 = mv20/2 + 3mg

At highest point: E = KE + mgy = 0 + 4mg

24
At A: E = mv20/2 + mgy0 = mv20/2 + 3mg

At highest point: E = KE + mgy = 0 + 4mg

So, E = mv20/2 + 3mg = 4mg

mv20/2 = mg
!
v0 = 2g = 4.43 m/s

What happens at B doesnt matter, provided there is no loss of energy


due to friction!

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Conservative Forces

Gravitational potential energy depends only on height

The difference in PE, mg(ho - hf) is independent of path taken

Gravity is a conservative force


26
Conservative Forces

Alternative definitions of conservative forces:

The work done by a conservative force in moving an object is


independent of the path taken.

A force is conservative when it does no net work in moving an


object around a closed path, ending up where it started.

In either case, the potential energy due to a conservative force


depends only on position.

Examples
Gravity
Elastic spring force
Electric force

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Non-conservative Forces

The work done by a non-conservative force depends on the path.

Friction the longer the path taken, the more the (negative)
work done by the friction force.

Defining a potential energy relies on a conservative force so that


the work done in moving an object from A to B depends only on
the positions of A and B.

Examples of non-conservative forces


Static and kinetic friction forces
Air resistance
Tension, or any applied force
Normal force
Propulsion force in a rocket

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