Question

State and prove law of conservation of energy in case of a freely falling body?

Answer 1

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 $K=\dfrac{1}{2}m{{v}^{2}}$.
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 ${{U}_{1}}=mgH$…..(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 ${{U}_{2}}=mgh$…….(2)
Hence the change in potential energy ($\Delta U$) is equal to ${{U}_{2}}-{{U}_{1}}$.
Let the speed of the ball be v. Hence, its kinetic energy will be ${{K}_{2}}=\dfrac{1}{2}m{{v}^{2}}$ and change in kinetic energy ($\Delta U$) will be ${{K}_{2}}=\dfrac{1}{2}m{{v}^{2}}$.
Let us calculate the kinetic energy of the ball at this height.
We know that F=ma and $a=v\dfrac{dv}{dx}$
Therefore, $F=m.v\dfrac{dv}{dx}$ .
$\Rightarrow F.dx=m.vdv$
Integrate both the sides.
$\Rightarrow \int{F.dx}=\int\limits_{{{v}_{1}}}^{{{v}_{2}}}{m.vdv}$ …….(i).
Here, F=mg and ${{v}_{1}}=0$, ${{v}_{2}}=v$.
Substitute the value of F, ${{v}_{1}}$, ${{v}_{2}}$ in equation (i).
$\Rightarrow \int{mg.dx}=\int\limits_{0}^{v}{m.vdv}$
$\Rightarrow mgx=\dfrac{1}{2}m{{v}^{2}}$ …… (ii).
x is the displacement of the ball. Hence, x=H-h.
Therefore, equation (ii) can be written as,
$mg(H-h)=\dfrac{1}{2}m{{v}^{2}}=\Delta K$ …..(iv)
From equations (1) and (2) we get,
$\Delta U={{U}_{2}}-{{U}_{1}}=mgh-mgH=-mg(H-h)$ ……(v)
Compare (iv) and (v).
We get that, $\Delta K=-\Delta U$.
We know that E=K+E.
$\Rightarrow \Delta E=\Delta K+\Delta U$.
But we found that $\Delta K=-\Delta U$.
Hence, $\Delta E=-\Delta U+\Delta U=0$
This means that the change in mechanical energy is zero. Therefore, the mechanical energy is constant or conserved.
Hence proved.

Note: We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. $W=\Delta K$.
We know that the change in potential energy is equal to the negative of work done on the body i.e. $W=-\Delta U$
Therefore, $\Delta K=-\Delta U$
And $\Delta E=\Delta K+\Delta U$.
Therefore, $\Delta E=-\Delta U+\Delta U=0$
Hence the mechanical energy of the body is conserved.