# A block of mass 100 g is moved with a speed of 5 m/s at the highest point in a closed circular...

## Question:

A block of mass 100 g is moved with a speed of 5 m/s at the highest point in a closed circular tube of radius 10 cm in a vertical plane. The cross-section of the tube is such that the block just fits in it. The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block during the process.

## Kinetic Energy:

The kinetic energy is defined as the energy which is composed in the material by virtue of its motion. In the standard international system, kinetic energy is measured in the units of a joule.

Given Data

• The mass of the block is {eq}m = 100\,{\rm{g}} = 0.1\,{\rm{kg}} {/eq}.
• The speed of the block is {eq}v = 5\,{\rm{m/s}}{/eq},
• The radius of the tube is {eq}r = 10\,{\rm{cm}} = 0.1\,{\rm{m}}{/eq}.

The expression of initial kinetic energy is calculated as.

{eq}{E_i} = \dfrac{1}{2}m{v^2}{/eq}

Substitute the value in the above expression.

{eq}\begin{align*} {E_i} &= \dfrac{1}{2}\left( {0.1\,{\rm{kg}}} \right){\left( {5\,{\rm{m/s}}} \right)^2}\\ {E_i}& = 1.25\,{\rm{J}} \end{align*}{/eq}

The final velocity of the block is zero.

The expression of final kinetic energy is calculated as.

{eq}{E_f} = \dfrac{1}{2}m{v^2}{/eq}

Substitute the value in the above expression.

{eq}\begin{align*} {E_i}& = \dfrac{1}{2}\left( {0.1\,{\rm{kg}}} \right){\left( 0 \right)^2}\\ {E_i} &= 0\,{\rm{J}} \end{align*}{/eq}

The total kinetic energy is calculated as.

{eq}\begin{align*} \Delta E &= {E_f} - {E_i}\\ \Delta E &= 0 - \left( {1.25\,{\rm{J}}} \right)\\ \Delta E &= - 1.25\,{\rm{J}} \end{align*}{/eq}

The work done due to gravitation is calculated as.

{eq}\begin{align*} {W_g} &= mg2r\\ {W_g}& = \left( {0.1\,{\rm{g}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)2\left( {0.1\,{\rm{m}}} \right)\\ {W_g} &= 0.196\,{\rm{J}} \end{align*}{/eq}

Here, {eq}g{/eq} is the acceleration due to gravity.

The expression of work energy theorem is given by,

{eq}{W_g} - {W_t} = \Delta E{/eq}

Substitute the value in the above expression.

{eq}\begin{align*} \left( {0.196\,{\rm{J}}} \right) - {W_t}& = \left( { - 1.25\,{\rm{J}}} \right)\\ {W_t} &= 1.25\,{\rm{J}} + 0.196\,{\rm{J}}\\ {W_t}& = 1{\rm{.446}}\,{\rm{J}} \end{align*}{/eq}

Thus, the work done by the tube on the block is {eq}{\rm{1}}{\rm{.446}}\,{\rm{J}}{/eq}.