Conservation of Mechanical Energy

From the previous section we know that the work done by a conservative force in terms of the change in potential energy is given by

ΔU = -WC(8.12)

where U is the potential energy and WCis the work done by a conservative force.

From the work-energy theorem, we know that

Wnet= ΔK

Where Wnetrepresents the sum of work done by all the forces acting on the mass.

If a particle is subject to only conservative forces, then

WC= Wnet= ΔK

Thus, the equation (8.12) becomes,

ΔU = -ΔK

orΔU + ΔK = 0 (8.14)

The equation (8.14) tells us that the total change in potential energy plus the total change in kinetic energy is zero if only conservative forces are acting on the system.

That is, there is no change in the sum of K plus U.

∴Δ(K + U) = 0 (8.15)

orΔE = 0where E = K + U

The quantity E = K + U is called the total mechanical energy.

According to equation (8.15), when only conservative forces act, the change in total mechanical energy of a system is zero, in otherwords,

If only conservative forces perform work on and within a system of masses, the total mechanical energy of the system is conserved.

Alternatively, the equation (8.15) may be written as

(Kf+ Uf) - (Ki+ Ui) = 0

orKf+ Uf= Ki+ Ui(8.16)

Since ΔE = 0, integrating both sides,

we getE = constant.

Problem Solving Strategy

1.Draw diagrams of the system showing initial and final configuration and assume a coordinate system.

2.Specify the reference level for potential energy. In case of spring, it is advisable to assume zero potential energy at the natural length of the spring. In case of gravity, any convenient level can be chosen as reference frame.

3.Looking at the initial configuration, ask yourself

What forms of energy are present?

• if the particle is moving include
• if the particle is not located at the reference level, include mgyi
• if the spring is stretched or compressed, include
• Looking at the final configuration, ask yourself

What forms of energy are present?

• if the particle is moving include
• if the particle is not located at the reference , include mgyf
• if the spring is stretched or compressed, include

Equate the initial and final total energies

Ki+ Ui = Kf+ Uf

Solve for the unknown.

Solution

The initial and final configurations are shown in the figure(8.22 b&c).

It is convenient to set Ug= 0 at the floor. Initially, only m2has potential energy. As it falls, it loses potential energy and gains kinetic energy. At the same time, m1gains potential energy and kinetic energy. Just before m ₂ lands, it has only kinetic energy. Let v the final speed of each mass. Then, using the law of conservation of mechanical energy.

Kf + Uf= Ki+ Ui

1/2(m ₁ + m ₂ )v ₂ + m ₁ gh= 0 + m ₂ gh

v ₂ =

Putting m1 = 3 kg; m ₂ = 5 kg; h = 5 m and g = 10 m/s ²

We get v ₂ =

orv = 5 m/s.

Example 8.10

Solution

putting l = 0.8 m; lo = 0.5 m; g = 10 m/s2, we get

v = √5 m/s

orv = 2.23 m/s

Example 8.11

Solution

To use Ef= Eiwe would need to assign the initial heights of the blocks arbitrary values h1 and h2. The corresponding potential energies, m1gh1 and m2gh2 would appear in both Ei and Efand hence would cancel.

We avoid this process by using the form ΔK + ΔU = 0 instead, since it does not require
U = 0 reference level.

(a)At the maximum extension xmax, the blocks come to rest, and thus ΔK = 0. Next, we must find the changes in Ugand US. When m2 falls by xmax, the spring extends by xmaxand m1 rises by xmaxsinθ. Therefore,

ΔK + ΔUg + ΔUS = 0

0 + (-m ₂ gx _max + m1gx _max sinθ) + 1/2kx ² max = 0

Thus,xmax= 2g/k(m2 - m1 sin θ) =   0.98 m

(b)In this case the change in kinetic energy is ΔK = 1/2(m1 + m2)v2.

The change in potential energy has the same form as in part (a), but with xmaxreplaced by
x = 0.5 m,.

= 0

Putting the given values we find v = 1.39 m/s.

Example 8.12

Solution

Let v be the velocity of the particle at the highest point.

According to Newton’s Second Law, the net force

Fnet= N + mg provides the centripetal force.

Therefore, N + mg = mv ² /R

The body is not detached from the loop if N ≥ 0. In the limiting case, N = 0.

That is mg = mv ² /R or v ₂ = gR

Applying energy conservation at the initial and highest point of the loop, we get

mgH = mg(2R) + 1/2mv ²

Usingv ₂ = gR, we obtain,

mgH = mg(2R) + 1/2m(gR)

orH = 5/2R = 2.5R

Putting R = 40 cm = 0.4 m, we get

H = (2.5) (0.4) = 1m.

Example 8.13

The bob of a simple pendulum of length L = 2 m has a mass m = 2 kg and a speed
v = 1 m/s when the string is at 35o to the vertical. Find the tension in the string at

(a)the lowest point in its swing

(b)the highest point

Solution

ΣFy = T – mg cosθ = mv ² /L(i)

To find the tension we need the speed, which can be found from the conservation law. We set Ug = 0 at the lowest point.

Note that the height is y = L – L cosθ. The mechanical energy is

E = (ii)

= 1/2 (2kg)(1 m/s) ² + (2kg)(9.8 N/kg)(2m)(1 – 0.8) = 9 J

(a)At the lowest point θ = 0; hence (ii) becomes

E = 1/2 mv ² max+ 0

Since E = 9 J we find vmax= 3 m/s.

Now we have the speed we can find the tension at θ = 0. From (i), T – mg = mv ² max/L, from which we find

T = 20 + 9 = 29 N

(b)At the highest point v = 0; hence, (ii) becomes

E = 0 + mgL(1 – cosθmax)

Using E = 9 J lead to cosθmax= 31/40

Sincev = 0, therefore, T = mgcosθmax = 15.5 N

Solution

We could set Ug= 0 at x = 0, but if we use the lowest point instead, all subsequent values are positive. Again the total energy has three terms E = K + Ug+ US.

(a)We set xo= 0.2 m and d = 0.5 m. Both Kiand Kfare zero, so Ei= and Ef= mgdsinθ. From work-energy theorem,

Ef– Ei= Wnc

mgdsinθ - = -fd

Thus f = 0.82 N

(b)The initial energy Eiis the same as above, but the final value at x = 0 is

Ef =

From work-energy theorem,

Ef– Ei= Wnc

1/2 mv ₂ + mgxosinθ - 1/2 kxo ₂ = - fxo

After solving we get, v = 2.45 m/s.