Question

Prove that the total mechanical energy of a body sliding down a smooth inclined plane under gravity remains constant ?

Answer 1

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The body of mass $\sf{m}$ moves along the inclined plane with an acceleration $\sf{a=g\sin\theta.}$

From the fig. we get relation between altitude of wedge and length of inclined surface as,

$\longrightarrow\sf{h=s\sin\theta\quad\quad\dots(i)}$

$\longrightarrow\sf{s=\dfrac{h}{\sin\theta}\quad\quad\dots(ii)}$

At point A, body has no velocity $\sf{(v_A=0),}$ hence its kinetic energy is zero.

• $\sf{K_A=0}$

The potential energy of the body at A is,

• $\sf{U_A=mgh}$

Hence total energy at A will be,

$\longrightarrow\sf{E_A=K_A+U_A}$

$\longrightarrow\sf{E_A=0+mgh}$

$\longrightarrow\sf{E_A=mgh\quad\quad\dots(1)}$

At point B, the body has velocity $\sf{v_B}$ given by third equation of motion as,

$\longrightarrow\sf{(v_B)^2=(v_A)^2+2gx\sin\theta}$

$\longrightarrow\sf{(v_B)^2=0^2+2gx\sin\theta}$

$\longrightarrow\sf{(v_B)^2=2gx\sin\theta}$

Hence kinetic energy of the body at B is,

• $\sf{K_B=\dfrac{1}{2}m(v_B)^2}$

• $\sf{K_B=\dfrac{1}{2}m(2gx\sin\theta)}$

• $\sf{K_B=mgx\sin\theta}$

When the body is at B, its height from ground will be $\sf{(s-x)\sin\theta.}$

Potential energy of body at B is,

• $\sf{U_B=mg(s-x)\sin\theta}$

From (ii),

• $\sf{U_B=mg\left(\dfrac{h}{\sin\theta}-x\right)\sin\theta}$

• $\sf{U_B=mg(h-x\sin\theta)}$

Hence total energy of the body at B will be,

$\longrightarrow\sf{E_B=K_B+U_B}$

$\longrightarrow\sf{E_B=mgx\sin\theta+mg(h-x\sin\theta)}$

$\longrightarrow\sf{E_B=mgx\sin\theta+mgh-mgx\sin\theta}$

$\longrightarrow\sf{E_B=mgh\quad\quad\dots(2)}$

At point C, the body has velocity $\sf{v_C}$ which is given by third equation of motion as,

$\longrightarrow\sf{(v_C)^2=(v_A)^2+2gs\sin\theta}$

$\longrightarrow\sf{(v_C)^2=0^2+2gs\sin\theta}$

$\longrightarrow\sf{(v_C)^2=2gs\sin\theta}$

So kinetic energy of body at C is,

• $\sf{K_C=\dfrac{1}{2}m(v_C)^2}$

• $\sf{K_C=\dfrac{1}{2}m(2gs\sin\theta)}$

• $\sf{K_C=mgs\sin\theta}$

From (i),

• $\sf{K_C=mgh}$

At C the body reaches the ground, hence it has no potential energy there.

• $\sf{U_C=0}$

Hence total energy of body at C is,

$\longrightarrow\sf{E_C=K_C+U_C}$

$\longrightarrow\sf{E_C=mgh+0}$

$\longrightarrow\sf{E_C=mgh\quad\quad\dots(3)}$

From (1), (2) and (3),

$\longrightarrow\sf{\underline{\underline{E_A=E_B=E_C}}}$

This implies total mechanical energy of the body remains constant throughout its motion, and that energy is converted from one form into another during its motion.

Hence the Proof!