Question

A 36.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m the floor is frictionless, and for the next 10.0 m the coefficient of friction is 0.20. What is the final speed of the crate after being pulled these 21.0 m?

Short answer

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

Step-by-step solution

Step 1:Understanding velocity and speed

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 information

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 \({v_{\rm{i}}} = 0\).

Calculate the expression for the net work done on the crate due to the change in the position of the crate

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:

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

Calculate the expression for the net work done on the crate due to the pull force and frictional force

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

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

Calculate the final speed of the crate

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 \(14.9\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).