Question

Explain Principle of Conservation of Energy with the help of examples :-
1) Free Fall of a body under Gravity
2) Oscillation of Spring-mass System
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Answer 1

Answer:

Hint: When a body falls freely, only the gravitational force influences its motion. During the fall the body possesses two types of energies, potential energy and kinetic energy. The sum of these energies is called mechanical energy.

Complete step by step answer:

Law of conservation of energy states that the energy of a system is always constant. In other words, we can say that energy can neither be created nor destroyed.

In the case of a freely falling body, it is the mechanical energy of the system that is conserved.

Mechanical energy (E) is the sum of the potential energy (U) and the kinetic energy (K) of the freely falling body. Therefore, E=K+U=constant.

Potential energy is defined as negative of the work done by the force affecting the body. In the case of gravity of earth, if a body of mass m is at height of h then its potential energy is given as U=mgh.

Kinetic energy is the energy possessed by a body when it is in motion. If a body of mass m is in motion with speed v, then its kinetic energy is equal to K=12mv2.

Let us now prove the conservation of the mechanical energy during a body falling freely.

When a body is falling feely, the only force affecting the motion of the body is the force of gravity exerted by earth, which is equal to F = mg.

Suppose we drop a ball of mass m from height of H.

The potential energy of the ball at height H is U1=mgH…..(1)

The kinetic energy of the ball at this height is zero because its speed is zero.

Then it will accelerate with the acceleration due to gravity (g).

Suppose that some time t the ball is at the height h. The potential energy of the ball at this point will be U2=mgh…….(2)

Hence the change in potential energy (ΔU) is equal to U2−U1.

Let the speed of the ball be v. Hence, its kinetic energy will be K2=12mv2 and change in kinetic energy (ΔU) will be K2=12mv2.

Let us calculate the kinetic energy of the ball at this height.

We know that F=ma and a=vdvdx

Therefore, F=m.vdvdx .

⇒F.dx=m.vdv

Integrate both the sides.

⇒∫F.dx=∫v1v2m.vdv …….(i).

Here, F=mg and v1=0, v2=v.

Substitute the value of F, v1, v2 in equation (i).

⇒∫mg.dx=∫0vm.vdv

⇒mgx=12mv2 …… (ii).

x is the displacement of the ball. Hence, x=H-h.

Therefore, equation (ii) can be written as,

mg(H−h)=12mv2=ΔK …..(iv)

From equations (1) and (2) we get,

ΔU=U2−U1=mgh−mgH=−mg(H−h)

Answer 2

★ Concept

According to this principle, energy can neither be created nor be destroyed, but can only be converted from one form to another, the total amount of energy of the universe remaining constant. In other words, whenever energy disappears in one form, an equal amount of energy appears in some other form. No violation of this principle has yet been observed.

Let's understand Principle of Conservation of energy with the help of some examples :-

⒈Free Fall of a body under Gravity

Let us consider the conservative of mechanical (kinetic + potential) energy during free-fall under gravity. Suppose a body of mass 'm' held at A at a height 'h' above the ground is allowed to fall freely under gravity. In the beginning, the body is at rest at A so that its kinetic energy is zero. It has only potential energy mgh, where g is acceleration due to gravity.

∴ Total energy of the body at A = K.E. + P.E. = 0 + mgh = mgh

Now, suppose during free-fall, when the body is passing across a point B at a distance x vertically below A, its velocity is v. Then,

v² = u² + 2gx = (0)² + 2gx = 2gx

∴ K.E. of the body at the Point B is

$\sf \: \dfrac{1}{2} m {v}^{2} = \dfrac{1}{2} m(2gx) = \bf \red{mgx}$

The P.E. of the body at the point B is mg(h-x) , because now the height of the body above the ground is (h-x).

∴ Total energy of the body at B = K.E. + P.E.= mgx + mg(h-x) = mgh

Now, suppose the body has fallen upto the point C and it just strikes the ground. Now its potential energy has become 'zero' , it has only kinetic energy.

The body has fallen through a height 'h'. If 'v' be the velocity of the body at C, then

v'² = u² + 2gh = (0)² + 2gh = 2gh

Therefore, the kinetic energy of the body is

$\sf \: \dfrac{1}{2} m {v'}^{2} = \dfrac{1}{2} m(2gh) = \bf \red{mgh}$

∴ Total energy of the body at C = K.E. + P.E.

= mgh + 0 = mgh.

Thus, the mechanical energy of a freely falling body remains constant at all positions.

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⒉ Oscillations of Spring mass system

We consider the oscillations of body of mass 'm' connected to a spring of force constant K . The equilibrium position of the body is A. The body is displaced to right to position B. At B, the potential energy of the stretched spring is 1/2 kx² , while the kinetic energy of the block, which is at rest is 0.

When released the block will move back to its equilibrium position A due to the elasticity of the spring. At A, where x = 0 , the potential energy of the unstretched spring becomes 0, being converted into the kinetic energy of the block which is now 1/2 kx².

Hence the block does not stop at A and goes to another extreme position C , where AC = -x . At C, the kinetic energy of the block becomes 0 being fully converted into the potential energy of the spring. The process is then repeated.

At all intermediate positions the energy of the spring block system is partly kinetic and partly potential, but the sum of the two is 1/2 kx² at all points.

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