A small block with mass \(0.0400\,{\rm{kg}}\)slides in a vertical circle of radius\(R = 0.500\,{\rm{m}}\)on the inside of a circular track. During one of the revolution of the block, when the block is the bottom of its path, point A, the normal force extended on the block by the track has magnitude\(3.95\,{\rm{N}}\). In this same revolution, when the block reaches the top of its path, point B, the normal force exerted on the block has magnitude\(0.680\,{\rm{N}}\). How much work is done on the block by friction during the motion of the block from point A to point B?

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 \( - 0.229\,{\rm{J}}\).