The law of Conservation of Mechanical Energy

Statement

The law of Conservation of Mechanical Energy states that if no external forces act (or the work done by them is zero) and the internal forces are conservative, the mechanical energy of the system remains constant.

What kind of question can we solve, and what kind of question can we not, using the law of Conservation of Mechanical Energy?

If non-conservative forces do work on the system or external forces do work on the system, we cannot use the conservation principle of total mechanical energy.

Suppose work done by non-conservative forces equals zero. In that case, work done by external forces equals zero, and internal forces operating within the system are conservative; then and only can we use the principle of conservation of total mechanical energy.

 

 

The law of Conservation of Mechanical Energy: Evidence through examples

Example 1: Freely falling object near the earth’s surface

Suppose a body of mass m is falling from rest near the earth’s surface. Initially (e = 0), it was at point A, at a height H above the earth’s surface.

Considering (earth + body) as a system and ground surface is considered a reference line. When the body is at point A, the total mechanical energy of the system is

            {(TM{E_{sys}})_A} = K{E_A} + P{E_A} = \frac{1}{2}m{(0)^2} + mgH = mgH

But at time t, the body reaches point B, the kinetic energy of the body increases, and the system’s potential energy decreases (because h < H). So, we see that the gain in the body’s kinetic energy is equal to the loss in potential energy of the system.

{(TM{E_{sys}})_B} = \frac{1}{2}mv_B^2 + mgh

According to the principle of conservation of energy :

{(TM{E_{sys}})_B} = {(TM{E_{sys}})_A}

\Rightarrow \quad \frac{1}{2}mv_B^2 + mgh = 0 + mgH

\Rightarrow \quad \underbrace {\frac{1}{2}mv_B^2 - 0}_{{\rm{gain in K}}{\rm{.E}}{\rm{.}}} = \underbrace {mgH - mgh}_{{\rm{loss in P}}{\rm{.E}}{\rm{.}}}

Example 2: A mass attached to a spring oscillating on a smooth surface

Consider a body moving with speed v0 on a smooth horizontal surface. As the body of mass m moves further, spring compresses. We are inserted to calculate the maximum compression produced in the spring.

Assuming block of mass m and spring as a system.

{(TM{E_{sys}})_{in}} = {(K.E.)_{Body}} + (P.E{._{sys}}) = \frac{1}{2}mv_0^2 + \frac{1}{2}k{(0)^2}

At the moment of maximum compression, the body’s speed becomes zero. Hence we see that the K.E. of the body decreases and the P.E. of the system increases.

So, loss in K.E. of the body is equal to gain in elastic potential energy.

{(TM{E_{sys}})_f} = {(K.E.)_{Body}} + (P.E{._{sys}}) = \frac{1}{2}m{(0)^2} + \frac{1}{2}kx_{\max }^2

According to the principle of conservation of TME.

{(TM{E_{sys}})_f} = {(TM{E_{sys}})_i}

\Rightarrow \quad \frac{1}{2}m{(0)^2} + \frac{1}{2}kx_{\max }^2 = \frac{1}{2}mv_0^2 + \frac{1}{2}k{(0)^2}

\Rightarrow \quad \underbrace {\frac{1}{2}kx_{\max }^2 - 0}_{{\rm{gain in P}}{\rm{.E}}{\rm{. of system}}} = \underbrace {\frac{1}{2}mv_0^2 - 0}_{{\rm{loss in K}}{\rm{.E}}{\rm{. of body}}}

 

The law of Conservation of Mechanical Energy: Equation

According to the work-energy theorem, as we know, the work done by all the forces equals the change in the kinetic energy.

Hence we can write,

{W_{{\rm{conservative}}}} = {W_{{\rm{non - conservative}}}} + {W_{{\rm{external}}}} = {K_f} - {K_i}            …(1)

Here, the three terms on the left denote the work done by the conservative internal forces, non-conservative internal forces, and external forces.

As we know, {W_{\rm{c}}} =  - ({U_f} - {U_i})        …(2)

Here, U stands for the potential energy of the system.

Now from (1) & (2),

{W_{{\rm{int}}}} + {W_{{\rm{ext}}}} = ({K_f} - {U_f}) - ({K_i} - {U_i})

{W_{{\rm{int}}}} + {W_{{\rm{ext}}}} = {(T.M.E.)_f} - {(T.M.E.)_i}    …(3)

Where T.M.E. = K + U is the total mechanical energy.

If the internal forces are conservative but external forces act on the system and they do work, {W_{{\rm{int}}}} = 0 then from equation (3) we can write,

{W_{{\rm{ext}}}} = {(T.M.E.)_f} - {(T.M.E.)_i}         …(4)

Therefore, we can write that the work done by external forces equals the change in the total mechanical energy of the system.

Here equation (4) is a mathematical form of energy conservation.

 

 

The Law of Conservation of Mechanical Energy: Solved Problems

Example 1: Finding work done by air friction

A body dropped from height h reaches the ground with speed 1.2\sqrt {gh}. Calculate the work done by air friction.

In above question {W_{{\mathop{\rmint}} }} \ne 0. Air friction is a non-conservative force. So, we cannot use the principle of conservation of total mechanical energy.

In this condition, we should use, work-energy theorem,

{(K.E.)_f} - {(K.E.)_i} = {W_{{\rm{all forces}}}}

Example 2: Finding spring constant

A block of mass m moving at a speed v compresses a spring through a distance x before its speed is halved. Find the spring constant of the spring.

Let the system is (spring + block)

Here, {W_{{\rm{non - conservative force}}}} = 0\,;\,({W_f} = 0)

{W_{{\rm{external force}}}} = 0\,;\,({W_{mg}} = 0,\,{W_N} = 0)

Internal force (spring force) is conservative.

So, we must use the principle of conservation of total mechanical energy.

{(T.M.E{._{sys}})_f} - {(T.M.E{._{sys}})_i} = {W_{ext}} + {W_{{\mathop{\rmint}} }}

{(T.M.E{._{sys}})_f} = {(T.M.E{._{sys}})_i}

Because, {W_{ext}} = 0\,;\,\,{W_{{\mathop{\rmint}} }} = 0 in above example.

Example 3: Freely falling object, Speed and height calculations

A particle is released from height H. At a certain height from the ground, its kinetic energy is twice its gravitational potential energy. Find the speed of particles and height at that moment.

Ans: Considering particles and earth as a system;

Here ;{W_{ext}} = 0\,;\,\,{W_{{\mathop{\rmint}} }} = 0 and internal forces operating within the system are conservative in nature.

K.E{._i} + P.E{._i} = K.E{._f} + P.E{._f}

\Rightarrow \quad \frac{1}{2}m{(0)^2} + mgH = K.E{._f} + mgh

\Rightarrow \quad \frac{1}{2}mv_f^2 = mg(H - h)

\Rightarrow \quad \frac{1}{2}mv_f^2 = mg\left( {H - \frac{H}{3}} \right)

\Rightarrow \quad v_f^2 = 2g \cdot \frac{{2H}}{3} = \frac{{4gH}}{3}

\Rightarrow \quad {v_f} = 2\sqrt {\frac{{gH}}{3}}

Also, K.E{._f} = 2{(P.E.)_f}

\Rightarrow \quad \frac{1}{2}mv_f^2 = 2mgh

\Rightarrow \quad 2mgh = mg(H - h)

\Rightarrow \quad 3mgh = mgH

\Rightarrow \quad h = \frac{H}{3}

Example 4: Finding work done using the conservation of energy approach

Under the action of force, 2 kg body moves such that its position x varies as a function of time given by x = \frac{{{t^3}}}{3}; x is in meter, t is in second. Calculate the work done by the force in the first two seconds.

Ans: Considering 2 kg body and earth as a system. Here, external force does work on the system, {W_{ext}} \ne 0.

Hence, we cannot use the energy conservation principle for this numerical.

So, here we will use work energy theorem,

K.E{._f} - K.E{._i} = {W_{ext}}        ({W_{mg}} = 0\,;\,{W_N} = 0)

\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = {W_{ext}}

\frac{m}{2}(v_f^2 - v_i^2) = {W_{ext}}       …(1)

Here, \frac{m}{2}(v_f^2 - v_i^2) = {W_{ext}}

\Rightarrow \quad v = {t^2}

{v_f} = {(2)^2} = 4\,{\rm{m/s}}

{v_i} = {(0)^2} = 0\,{\rm{m/s}}

Now, from (1),

\frac{2}{2}\left( {{4^2} - {0^2}} \right) = {W_{ext}}

\Rightarrow \quad {W_{ext}} = 16\,J

 

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