91Ó°ÊÓ

Newton's second law is a mathematical way to describe how a force can change the way a body moves. It says that the rate at which a body's momentum changes over time is equal in magnitude and direction to the force that acts on it.

Here we have, the mass of block is\(m = 0.04\,{\rm{kg}}\).

The radius of circle is \(R = 0.5\,{\rm{m}}\).

The normal force on the block at the bottom of the circle is \({n_A} = 3.95\,{\rm{N}}\) and the normal force on the block at the top of the circle is \({n_B} = 0.68\,{\rm{N}}\)

Here we have other forces than gravity is acting on the block.

So, work-energy theorem is given by:

\({K_1} + {U_1} + {W_{other}} = {K_2} + {U_2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 1 \right)\)

Where kinetic energy is given by:

\(K = \frac{1}{2}m{v^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 2 \right)\)

And the gravitational potential energy is given by:

\({U_{grav}} = mgy\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 3 \right)\)

We also know that, in a uniform circular motion, the acceleration is directed towards the center of the circle and its magnitude is given by:

\({a_{rad}} = \frac{{{v^2}}}{R}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 4 \right)\)

Now, apply Newton’s second law to the block at the bottom of the circle along the radial direction.

Therefore, we get

\(\begin{aligned}{}\sum {{F_{rad}} = {n_A} - mg = m{a_{rad}}} \\ \Rightarrow {a_{rad}} = \frac{{{n_A}}}{m} - g\end{aligned}\)

Put value of \({a_{rad}}\) in equation (4),

\(\begin{aligned}{}\frac{{{v_A}^2}}{R} = \frac{{{n_A}}}{m} - g\\ \Rightarrow {v_A} = \sqrt {\frac{{{n_A}R}}{m} - gR} \\ \Rightarrow {v_A} = \sqrt {\frac{{3.95 \times 0.5}}{{0.04}} - 9.8 \times 0.5} = 6.67\,{\rm{m/s}}\end{aligned}\)

Now, apply Newton’s second law to the block at the top of the circle along the radial direction.

Therefore, we get

\(\begin{aligned}{}\sum {{F_{rad}} = {n_B} + mg = m{a_{rad}}} \\ \Rightarrow {a_{rad}} = \frac{{{n_A}}}{m} + g\end{aligned}\)

Put value of \({a_{rad}}\) in equation (4),

\(\begin{aligned}{}\frac{{{v_B}^2}}{R} = \frac{{{n_B}}}{m} + g\\{v_B} = \sqrt {\frac{{{n_B}R}}{m} + gR} \\{v_B} = \sqrt {\frac{{0.68 \times 0.5}}{{0.04}} + 9.8 \times 0.5} = 3.66\,{\rm{m/s}}\end{aligned}\)

Let\(y = 0\)at the bottom of the circle.

Now for point A

\({y_A} = 0,\;\;\;\;{v_A} = 6.67\,{\rm{m/s}}\)

For point B,

\({y_B} = 2R = 1.00\,{\rm{m}},\;\;\;\;{v_B} = 3.66\,{\rm{m/s}}\)

Here block start from zero gravitational potential energy level.

Therefore, \({U_1} = 0\)

Now, put value of \(m\;{\rm{and }}{{\rm{v}}_1}\) into equation (2),

\( \Rightarrow {K_1} = \frac{1}{2} \times 0.04 \times {\left( {6.67} \right)^2} = 0.8895\,{\rm{J}}\)

Now, put value of \(m\;{\rm{and }}{{\rm{v}}_2}\) into equation (2),

\( \Rightarrow {K_2} = \frac{1}{2} \times 0.04 \times {\left( {3.66} \right)^2} = 0.268\,{\rm{J}}\)

Now, put value of \(m\;{\rm{and }}{{\rm{y}}_1}\) into equation (3),

\( \Rightarrow {U_2} = 0.04 \times 9.8 \times 1.00 = 0.392\,{\rm{J}}\)

Now, put these values in equation (1),

\(\begin{aligned}{} \Rightarrow 0.8895 + 0 + {W_f} = 0.268 + 0.392\\ \Rightarrow {W_f} = - 0.229\,{\rm{J}}\end{aligned}\)

Hence work is done on the block by friction during the motion of the block from point A to point B is \( - 0.229\,{\rm{J}}\).