Question

A factory worker moves a 30.0 kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25.

1. What magnitude of force must the worker apply?
2. How much work is done on the crate by the worker's push?
3. How much work is done on the crate by friction?
4. How much work is done by normal force? By gravity?
5. What is the net work done on the crate?

Answer 1

Concepts and reason

The concept used to solve this problem is the work-energy theorem.

Initially, the friction force can be calculated by multiplying the coefficient of friction and the normal force. Later, work done can be calculated by multiplying the force and the distance covered by the worker. Finally, net work done can be calculated by adding all the work done.

Fundamentals

The expression for the friction force is,

${F_f} = {\mu _k}mg$

Here, the friction force is ${F_f}$, the coefficient of the friction is ${\mu _k}$, the mass of the crate is $m$, and the gravitational acceleration is $g$.

The formula for the work done is:

$W = r{F_f}$

Here, the work done is $W$, and the distance covered is $r$.

The expression for the net work done is,

${W_{net}} = {W_f} + {W_w} + {W_N} + {W_g}$

Here, the net work done is ${W_{net}}$, the work done by friction force is ${W_f}$, the work done by the worker is ${W_w}$, the work done by normal force is ${W_N}$, and the work done by gravitation force is ${W_g}$.

(1)

The expression for the friction force is,

${F_f} = {\mu _k}mg$

Substitute $0.25$for${\mu _k}$, $30\,{\rm{kg}}$ for $m$, and $9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$for$g$.

$\begin{array}{c}\\{F_f} = \left( {0.25} \right)\left( {30\,{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\\\ = 73.5{\rm{N}}\\\end{array}$

(2)

The expression for the work done by the worker is,

${W_w} = r{F_f}$

Substitute $73.5{\rm{N}}$for${F_f}$ and $4.5\,{\rm{m}}$ for $r$

$\begin{array}{c}\\{W_w} = \left( {4.5\,{\rm{m}}} \right)\left( {73.5\,{\rm{N}}} \right)\\\\ = 330.75\,{\rm{J}}\\\end{array}$

(3)

The expression for the work done by the friction force is,

${W_f} = - r{F_f}$

Substitute $73.5{\rm{N}}$for${F_f}$ and 4.5m for $r$.

$\begin{array}{c}\\{W_f} = - \left( {\left( {4.5\,{\rm{m}}} \right)\left( {73.5\,{\rm{N}}} \right)} \right)\\\\ = - 330.75\,{\rm{J}}\\\\ \approx - 331\,{\rm{J}}\\\end{array}$

(4)

The work done by the normal force is,

$\begin{array}{c}\\{W_N} = N\left( r \right)\cos \left( {90} \right)\\\\ = 0\,{\rm{J}}\\\end{array}$

The work done by gravitational force is,

$\begin{array}{c}\\{W_g} = r{F_g}\cos \left( {90} \right)\\\\ = 0\,{\rm{J}}\\\end{array}$

(5)

The expression for the total work done is:

${W_{net}} = {W_f} + {W_w} + {W_N} + {W_g}$

Substitute $330.75\;{\rm{J}}$for${W_w}$,$- \;330.75\;{\rm{J}}$for ${W_f}$, ${\rm{0J}}$for${W_N}$ , and ${\rm{0}}\,{\rm{J}}$for${W_g}$

$\begin{array}{c}\\{W_{net}} = 330.75 - 330.75 + 0 + 0\\\\ = 0\;{\rm{J}}\\\end{array}$

Ans: Part 1

The magnitude of the force that the worker must apply is${\bf{73}}{\bf{.5}}\,{\bf{N}}$.

Part 2

The work done by the worker is${\bf{3}}{\bf{.31}}\,{\bf{kJ}}$.

Part 3

The work done by the friction is${\bf{ - 331}}\,{\bf{J}}$

Part 4

The work done by the gravitational force is ${\bf{0}}\,{\bf{J}}$ and work done by the normal forces is ${\bf{0}}\,{\bf{J}}$.

Part 5

The net work done by the system is${\bf{0}}\,{\bf{J}}$.