Work Power & Energy

The t - F diagram is a straight line passing through (0, 4) and (3, 0). The equation of this straight line is

$F = - {4 \over 3}t + 4$.

Newton's second law gives acceleration of the block as

$a = F/m = - {2 \over 3}t + 2$.

Integrate to get the velocity v at 4.5 s

$v = \int_0^{4.5} {a\,dt = \int_0^{4.5} {\left( { - {2 \over 3}t + 2} \right)dt = 2.25} } $ m/s.

Thus, kinetic energy of the block at 4.5 s is

$K = {1 \over 2}m{v^2} = 5.06$ J.

When the block B is displaced towards wall-1 only spring S$_1$ is compressed and S$_2$ is in natural state.

If spring S$_2$ is not attached to wall but it is free but it gain momentum ($\uparrow$) and moves towards S$_1$.

The spring S$_1$ comes to natural length and spring S$_2$ gets compressed.

So if there is no frictional force i.e.,

$fr=0$

So potential energy of spring S$_1$ as the potential energy of spring.

Hence, ${(PE)_{{S_1}}} = {(PE)_{{S_2}}}$

$ \Rightarrow {1 \over 2}{k_1}{x^2} = {1 \over 2}{k_2}y$

$ \Rightarrow 1k{x^2} = (4k){y^2} \Rightarrow {x^2} = 4{y^2}$

Potential energy of spring.

Hence, ${(PE)_{{S_1}}} = {(PE)_{{S_2}}}$

$ \Rightarrow {1 \over 2}{k_1}{x^2} = {1 \over 2}{k_2}{y^2}$

$ \Rightarrow {1 \over 2}k{x^2} = {1 \over 2}(4k){y^2}$

$ \Rightarrow {x^2} = 4{y^2}$

$ \Rightarrow {y \over x} = {1 \over 2}$

The bob of mass m requires velocity to reach at a point with vertical circular motion.

$V = \sqrt {5gl} $

Then, $h = L - L\cos \theta = L(1 - \cos \theta )$

Applying law of conservation of energy at point A & C

Kinetic energy (K.E)_A + Potential energy (U_A) = Kinetic energy (K.E_C) + Potential energy (U_C)

$ \Rightarrow {1 \over 2}m{V^2} + 0 = {1 \over 2}m{\left( {{{\sqrt {5gl} } \over 2}} \right)^2} + mgL(1 - \theta )$

$ \Rightarrow {{5L} \over 2} = {{5L} \over 8} + L(1 - \cos \theta )$ ($\because$ $V = \sqrt {5gl} $)

$ \Rightarrow {5 \over 2} - {5 \over 8} = 1 - \cos \theta $

$ \Rightarrow {{20 - 5} \over 8} = 1 - \cos \theta $

$ \Rightarrow \cos \theta = 1 - {{15} \over 8}$

$ \Rightarrow \cos \theta = {{ - 7} \over 8}$

$ \Rightarrow \theta = 152^\circ $

$\therefore$ ${{3\pi } \over 4} < \theta < \pi $

Case I:

Case II : Tilted at 30$^\circ$.

$N = mg\cos 30^\circ $

$ = mg{{\sqrt 3 } \over 2}$

${f_2} = \mu B = rmg{{\sqrt 3 } \over 2}$

${f_1} = {l \over N} = {l \over {mg{{\sqrt 3 } \over 2}}}$

Case III : $\Delta$E $ = {1 \over 2}m{v^2} - mgh$ = Decreases

Option (C) is correct because friction is depend on surface.

Case I :

Statement 1 : In the 1^st case the mechanical energy is completely converted into heat energy because of friction.

Case II :

M.E. is converted into heat due to friction but another part of M.E. is retained in the form of P.E. of block. Therefore statement I is correct.

Statement 2 : This is a wrong statement because the coefficient of friction between block and surface does not depend on angle of inclination.

The work done by the friction force is zero in rolling without slipping. The surface BC is frictionless. Thus mechanical energy of the balls is conserved.

i.e., it is same at point $\mathrm{A}, \mathrm{B}$ and C .

If $K_A, K_B$ and $K_C$ represent the total oenergies (transtational plus rotational)

Then, $\mathrm{K}_{\mathrm{A}}+\mathrm{mgh}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}=\mathrm{K}_{\mathrm{C}}+\mathrm{mgh}_{\mathrm{C}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

The equation (i) gives $K_B>K_C$ and $K_B>K_A$ irrespective of the relation between $\mathrm{h}_{\mathrm{A}}$ and $\mathrm{h}_{\mathrm{C}}$.

Let us analyses the problem a little more.

Let $K_B{ }^{\text {tans }}$ and $K_C{ }^{\text {trans }}$ are translational kinetics energies and $K_B{ }^{\text {rot }}$ and $K_C{ }^{\text {rot }}$ rotational kinetic energies of balls.

The equation (i) can be written as

$ \begin{aligned} \mathrm{K}_{\mathrm{A}}+m g h_A & =\left(\mathrm{K}_{\mathrm{B}}\right)_{\operatorname{tran}}+\left(\mathrm{K}_{\mathrm{B}}\right)_{\mathrm{rot}} \\ & =\left(\mathrm{K}_{\mathrm{C}}\right)_{\operatorname{tran}}+\left(\mathrm{K}_{\mathrm{C}}\right)_{\mathrm{rot}}+m g h_{\mathrm{C}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

Track $B C$ is frictionless and there is no torque on the ball about its centre of mass. Hence, its angular acceleration is zero, i.e., it moves with a constant angular velocity, which implies.

$ \left(\mathrm{K}_{\mathrm{B}}\right)_{\mathrm{rot}}=\left(\mathrm{K}_{\mathrm{C}}\right)_{\mathrm{rot}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

The equation (iii) and (iv) gives $\left(\mathrm{K}_{\mathrm{B}}\right)_{\text {trans }}>\left(\mathrm{K}_{\mathrm{C}}\right)_{\text {trans }}$.