CONSERVATION OF ENERGY

1. Potential energy is, as the word suggests, the energy “locked up” up somewhere and which
can do work. Potential energy kam karnay ki salahiat hai! Potential energy can be converted
into kinetic energy, 1 2
2 mv . As I showed you earlier, this follows directly from Newton’s
Laws.
2. If you lift a stone of mass from the ground up a distance , you have to do work against
gravity. The (constant) force is , and so . By conservation of energy,
the work done by you
m x
mg W = mgx
was transformed into gravitational potential energy whose
values is exactly equal to . Where is the energy stored? Answer: it is stored neither
in the mass or in the earth - it is stored in
mgx
the gravitational field of the combined system
of stone+earth.
3. Suppose you pull on a spring and stretch it by an amount away from its normal
(equilibrium) position. How much energy is sto
x
red in the spring? Obviously, the
spring gets harder and harder to pull as it becomes longer. When it is extended by
length x and you pull it a further distance dx, the small amount of work d
2
0 0
one is
. Adding up all the small bits of work gives the total work:
1
2
This is the work you did. Maybe you got tired working so
x x
dW Fdx kxdx
W Fdx k xdx k x
= =
=∫ =∫ =
1 2
2
hard. What was the result
of your working so hard? Answer: this work was transformed into energy stored in the
spring. The spring contains energy exactly equal to .
4. Kinetic energy obvi
k x
ously depends on the frame you choose to measure it in. If you are
running with a ball, it has zero kinetic energy with respect to you. But someone who is
standing will see that it has kinetic energy! Now consider the following situation: a box
of mass 12kg is pushed with a constant force so that so that its speed goes from zero to
1.5m/sec (as measured by the person at rest on the cart) and it covers a distance of 2.4m.
Assume there is no friction.
2.4m
v=1.5m/s
mass of box =12 kg
15m/s
h h
h / 2
A B
C
0 v
D
2
2 2 2
Let's first calculate the change in kinetic energy:
1 (12 )(1.5 / ) 0 13.5
2
And then the (constant) acceleration:
v v (1.5 / ) 0
2( )
f i
f i
f i
K K K kg m s J
a m s
x x
Δ = − = − =
− −
= =
−
2
2
0.469 /
2(2 4 )
This acceleration results from a constant net force given by:
(12 )(0.469 / ) 5.63
From this, the work done on the crate is:
(
m s
m
F ma kg m s N
W F x
=
⋅
= = =
= Δ = 5.63N)(2.4m)=13.5J (same as ΔK= 13.5J !)
5. Now suppose there is somebody standing on the ground, and that the trolley moves at
15 m/sec relative to the ground:
2 2
f i
2 2
Let us repeat the same calculation:
1 v 1 v
2 2
1(12 )(16.5 / ) 1(12 )(15.0 / ) 284
2 2
This example clearly shows that work and
f i K K K m m
kg m s kg m s J
Δ ′= ′ − ′= −
= − =
energy have different values in different frames.
6. The total mechanical energy is: . If there is no friction then is
conserved. This means that the sum does not change with time. For example: a ball
is thrown upwards at
mech mech E =KE+PE E
0
2
2 0
0
speed v . How high will it go before it stops? The loss of potential
energy is equal to the gain of potential energy. Hence, 1v v.
2 2
Now look at the smooth, frictionless motion of
m mgh h
g
= ⇒ =
a car over the hills below:
15m/s
15m/s 16.5m/s
50.4m
48.0m
2 2
Even though the motion is complicated, we can use the fact that the total energy is a
constant to get the speeds at the points B,C,D:
At point A: 1 v 1 v
2 A 2 B m +mgh= m +
2 2 2
C A
2 2 2
D D A
v v
At point C: 1 v 1 v v v
2 2 2
At point D: 1 v 1 v v v 2
2 2
7. Remember that pot
B C
A C
A
mgh
m mgh m mgh gh
m mgh m gh
⇒ =
+ = + ⇒ = +
+ = ⇒ = +
ential energy has meaning only for a force that is conservative. A
conservative force is that for which the work done in going from point A to point B
is independent of the path chosen. Frictio
2
n is an example of a non-conservative force
and a potential energy cannot be defined. For a conservative force, F = . So, for
a spring, 1 and so .
2
8. Derivation of F = :
dV
dx
V kx F kx
dV
dx
−
= =−
−
0
If the particle moves distance in a potential , then
change in PE is where, . From this, . Now let 0.
Hence, F = lim V = .
x
x V
V V F x F V x
x
dV
Δ → x dx
Δ
Δ
Δ Δ = − Δ = − Δ →
Δ
Δ
− −
Δ