JR/2008

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.

JR/2008

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

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

2gh v

JR/2008

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 = = 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

2gh v 10 x 9.81 x 2

JR/2008

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.

JR/2008

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 = ?

to find acceleration (retardation) a use

acceleration = = -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

Conservation of Energy

asuv 222 s

uva

2

22

015.02

140 22

JR/2008

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 !!!