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Velocity can be described as the rate at which the object shifts its location while moving from one location to another.

It is a vector quantity; so, it depends on direction. Speed is the distance covered in a unit of time.

It is a scalar quantity; so, it does not depend upon direction. It depends on magnitude only.

Given data:

The mass of the crate is \(m = 36.0\;{\rm{kg}}\).

The horizontal constant pull force is \(F = 225\;{\rm{N}}\).

The total distance moved by the crate is \(d = 21.0\;{\rm{m}}\).

The distance moved by the crate when frictional force is applied is \({d_{{\rm{fr}}}} = 10.0\;{\rm{m}}\).

The coefficient of friction is \(\mu = 0.20\).

It is given that the cart starts from rest. Therefore, its initial velocity will be \({v_{\rm{i}}} = 0\).

The work done on the crate to change its state is equal to the change in the kinetic energy. Therefore, the expression for the net work done can be written as:

\(\begin{aligned}{W_{{\rm{net}}}} &= \Delta KE\\{W_{{\rm{net}}}} &= \frac{1}{2}mv_{\rm{f}}^{\rm{2}} - \frac{1}{2}mv_{\rm{i}}^{\rm{2}}\\{W_{{\rm{net}}}} &= \frac{1}{2}mv_{\rm{f}}^{\rm{2}} - \frac{1}{2}m\left( 0 \right)\\{W_{{\rm{net}}}} &= \frac{1}{2}mv_{\rm{f}}^{\rm{2}}\end{aligned}\)鈥� (i)

The pull force and friction force are responsible for the work done. So, the net work done will be:

\(\begin{aligned}{W_{{\rm{net}}}} &= {W_{\rm{p}}} + {W_{{\rm{fr}}}}\\{W_{{\rm{net}}}} &= Fd\cos \left( {0^\circ } \right) + {F_{{\rm{fr}}}}{d_{{\rm{fr}}}}\cos \left( {180^\circ } \right)\\{W_{{\rm{net}}}} &= Fd - {\mu _{\rm{k}}}mg{d_{{\rm{fr}}}}\end{aligned}\)鈥� (ii)

Equate equations (i) and (ii) to get the value of the final speed of the crate.

\(\begin{aligned}\frac{1}{2}mv_{\rm{f}}^{\rm{2}} &= Fd - {\mu _{\rm{k}}}mg{d_{{\rm{fr}}}}\\{v_{\rm{f}}} &= \sqrt {\frac{2}{m}\left( {Fd - {\mu _{\rm{k}}}mg{d_{{\rm{fr}}}}} \right)} \\{v_{\rm{f}}} &= \sqrt {\frac{2}{{\left( {36.0\;{\rm{kg}}} \right)}}\left( {\left\{ {\left( {225\;{\rm{N}}} \right)\left( {21.0\;{\rm{m}}} \right)} \right\} - \left\{ {\left( {0.20} \right)\left( {36.0\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {10.0\;{\rm{m}}} \right)} \right\}} \right)} \\{v_{\rm{f}}} &= 14.9\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the final speed of the crate is \(14.9\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).