Question

An 8 kg package in a mail-sorting room slides 2 m down a chute that is inclined at 53 degrees below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.4.

Calculate the work done on the package by...
A) Friction
B) Gravity
C) The normal force
D) What is the net work done on the package?

Answer 1

Concepts and reason

The main concept used to solve the problem is relationship between work done, force and distance.

Initially, determine the forces acting on the block. Later, find the displacement of the block along the direction of force. Finally, multiply the force and displacement to calculate the work done.

Fundamentals

The work done on an object is,

$W = F \cdot d$

Here, $W$ is the work done, $F$ is the force, and $d$ distance.

(A)

The work done due to friction is,

${W_{friction}} = {F_{frictional}} \cdot d$

Here, ${W_{friction}}$ is the work done due to frictional force, ${F_{frictional}}$ is the frictional force, and $d$ is the displacement due to the frictional force.

The frictional force is,

${F_{frictional}} = - {\mu _k}N$

Here, ${\mu _k}$ is the coefficient of friction and $N$ is the normal force.

The negative sign indicates that the frictional force acts in the opposite direction of the block’s motion.

The normal force is,

$N = mg\cos \theta$

Here, $m$ is the mass and $g$is the acceleration due to gravity, and $\theta$ is the inclination.

Substitute $N = mg\cos \theta$ in ${F_{frictional}} = {\mu _k}N$ to calculate the frictional force. The frictional force is,

$\begin{array}{c}\\{F_{frictional}} = - {\mu _k}\left( {mg\cos \theta } \right)\\\\ = - {\mu _k}mg\cos \theta \\\end{array}$

Substitute ${F_{frictional}} = - {\mu _k}mg\cos \theta$ in ${W_{friction}} = {F_{frictional}} \cdot d$ to calculate the work done due to frictional force.

The work done is,

$\begin{array}{c}\\{W_{friction}} = \left( { - {\mu _k}mg\cos \theta } \right) \cdot d\\\\ = - {\mu _k}mgd\cos \theta \\\end{array}$

Substitute $0.4$ for ${\mu _k}$, $8.0\,{\rm{kg}}$ for $m$, $2.0\,{\rm{m}}$ for $d$, $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for $g$, and $53^\circ$ for $\theta$.

The work done is,

$\begin{array}{c}\\{W_{friction}} = - \left( {0.4} \right)\left( {8.0\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right)\left( {2.0\,{\rm{m}}} \right)\left( {\cos \left( {53^\circ } \right)} \right)\\\\ = - 37.7\,{\rm{J}}\\\end{array}$

The work done due to friction is $- 37.7\,{\rm{J}}$.

The negative sign indicates that the energy is lost due to work done by friction.

(B)

The work done due to gravity is,

${W_{gravity}} = {F_{gravitational}} \cdot d$

Here, ${W_{gravity}}$ is the work done due to gravitational force, ${F_{gravitational}}$ is the gravitational force and $d$ is the displacement due to the gravitational force.

The gravitational force is,

${F_{gravitational}} = mg\sin \theta$

Here, $m$ is the mass, $\theta$ is the inclination and $g$ is the acceleration due to gravity.

Substitute ${F_{gravitational}} = mg\sin \theta$ in ${W_{gravitational}} = {F_{gravitational}} \cdot d$ to calculate the work done due to gravitational force. The work done is,

$\begin{array}{c}\\{W_{gravitational}} = \left( {mg\sin \theta } \right) \cdot d\\\\ = mgd\sin \theta \\\end{array}$

Substitute $8.0\,{\rm{kg}}$ for $m$, $2.0\,{\rm{m}}$ for , $53^\circ$ for $\theta$, and $9.8\,{\rm{m/}}{{\rm{s}}^2}$ for $g$.

$\begin{array}{c}\\{W_{gravitational}} = \left( {8.0\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right)\left( {2.0\,{\rm{m}}} \right)\left( {\sin \left( {53^\circ } \right)} \right)\\\\ = `125.2\,{\rm{J}}\\\end{array}$

The work done due to gravity is $125.2\,{\rm{J}}$.

(C)

The work done due to the normal force is,

${W_{normal}} = N \cdot d$

Here, ${W_{normal}}$ is the work done due to normal force, $N$ is the normal force and $d$ is the displacement due to the normal force.

The displacement due to normal force is zero. Substitute $0$ for $d$. The work done is,

$\begin{array}{c}\\{W_{normal}} = N\left( 0 \right)\\\\ = 0\,{\rm{N}}\\\end{array}$

The work done due to normal force is $0\,{\rm{J}}$

(D)

The net work done is,

$W = {W_{frictional}} + {W_{gravitational}} + {W_{normal}}$

Substitute $- 37.7\,{\rm{J}}$ for ${W_{frictional}}$, $0\,{\rm{J}}$ for ${W_{normal}}$ and $125.2\,{\rm{J}}$ for ${W_{gravitational}}$. The net work done is,

$\begin{array}{c}\\W = - 37.7\,{\rm{J}} + 0\,{\rm{J}} + 125.2\\\\ = 87.5\,{\rm{J}}\\\end{array}$

The net work done on the block is $87.5\,{\rm{J}}$.

Ans: Part A

The work done due to frictional force is $- 37.7\,{\rm{J}}$.

Part B

The work done due to gravity is $1.2 \times {10^2}\,{\rm{J}}$.

Part C

The work done due to normal force is $0\,{\rm{J}}$

Part D

The net work done on the block is $87.5\,{\rm{J}}$.