Transcript Cons_of_Energy

Conservation of Energy
Conservation of Energy:- the principle states that:ENERGY CANNOT BE CREATED OR DESTROYED
This can be put to good use in the solution of engineering problems
A typical example can be seen when considering free falling objects
eg. If a smooth ball of mass m kg is held h metres above a fixed surface
and released it will fall due to gravity g.
In energy terms the initial phase is potential energy (PE) mgh
as the ball falls the PE is transferred into kinetic energy (KE) ½ mv2
If energy losses are ignored then the PE at the start = KE at the end
when the ball hits the surface.
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Conservation of Energy
If
PE at the start
= KE at the end
then
mgh
=
½ mv2
as mass m is constant the equation can be transposed for v final velocity
v  2gh
Students should recall this equation from previous work on free fall.
It must be noted that if energy losses are included this equation cannot
provide an accurate solution
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Conservation of Energy
From
PE at the start = KE at the end
and
mgh
=
½ mv2
The concept of energy to work translation can also be included
eg. If a smooth ball of mass 3 kg is held 10 metres above a fixed
absorbent surface and released it will fall due to gravity g.
Find the average retarding force provided by the surface on the ball.
Initial PE (mgh) = 3kg x 10m x 9.81 = 294.3 J
The velocity v on impact = v  2gh
 2 x 9.81 x 10
= 14 ms-1
But without losses the KE on impact = PE at start = 294.3 J
If the ball penetrates a depth of 15mm into the surface then work is done
As work = force x distance the 294.3 J of energy will be given up to work
As the distance moved into the surface = 15mm = 0.015m
Then the average retarding force on the ball can be found
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Conservation of Energy
From before the KE at impact = 294.3J.
(Note this was also the initial PE)
This is transferred into work (Force x distance) F.s
Thus 294.3 J = Average force F x 0.015m
Giving F = 294.3/0.015 = 19620 N = 19.62 kN
This will be the average force on the ball provided by the surface.
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Conservation of Energy
Alternative solution:
From the free fall equation it was noted that velocity at impact was 14 ms-1
Treating the problem as a simple motion task:
u = 14 ms-1
v = 0 ms-1 (when the ball finally stops 15mm into the surface)
a = ?
s = 15mm = 0.015m
t = ?
2
2
to find acceleration (retardation) a use
v  u  2as
acceleration =
0 2  14 2
2  0.015
v2  u 2
a
2s
= -6533.33 ms-2 (retardation)
To find the retarding force F use N2; F = ma = 3kg x 6533.33 ms-2 = 19.6 kN
Review both methods and consider which you feel to be the most efficient method
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Conservation of Energy
Try the following:
A bullet of mass 10 grams leaves a gun at a velocity of 350 ms-1. If the bullet
hits a stationary target which provides an average retarding force of 1234 N
calculate how far into the target the bullet will penetrate before stopping.
Neglect air resistance on the bullet.
Soln.
Initial KE = KE at impact (no losses) = ½ mv2
when mass m = 10 gram = 0.01kg, velocity v = 350 ms-1
KE = ½ x 0.01 x 3502 = 612.5J
work done on target = 612.5J
work = Force x distance (Fs) When retarding force F = 1234N
Distance moved = 612.5 J ÷ 1234N = 0.496metres = 500 mm penetration
Consider a bullet proof vest !!!
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