Question

A box weighing 2000 N is to be slowly slid through 20 m on a straight track having a friction coefficient 0.2 with the box.
(a) Find the amount of work done by the person pulling the box with a chain at an angle θ with the horizontal.
(b) Find the amount of work done when the person has chosen a value of θ which ensures him the minimum magnitude of the force

Answer 1

Hint: For just sliding the box without any acceleration Force applied on the box will be just enough to counterbalance the frictional force provided by the floor.
Apply the equilibrium conditions: $\sum {Fx = 0} {\text{ and }}\sum {Fy = 0} $
Then, find the work done by the expression:
Work done = component of force acting in the direction of displacement x total displacement of the box

Complete step by step answer:
Given, W =2000 N
$\mu = 0.2$
And displacement, d = 20m
Drawing free body diagram of the box:

Forces acting on the box are:
Weight of the box acting downwards, W = mg=2000N
Normal reaction of the floor R, acting upwards.
Components of force F i.e. \[F\cos \theta {\text{ and }}F\sin \theta \], acting in right and downward direction respectively.
Frictional force, $f = \mu R = 0.2R$ acting in the left direction.
So, for just sliding the box without any acceleration Force applied on the box will be just enough to counterbalance the frictional force provided by the floor.
Hence box will be in equilibrium, applying equilibrium conditions:
$\sum {Fy = 0} $(i.e. sum of all vertical forces should be zero):
$\begin{gathered}
   \Rightarrow R + F\sin \theta - W = 0 \\
   \Rightarrow R + F\sin \theta - 2000 = 0 \\
   \Rightarrow R = 2000 - F\sin \theta {\text{ - - - - - - - - - - - - - - - - - - eq}}{\text{.(i)}} \\
\end{gathered} $


Also, $\sum {Fx = 0} $ (i.e. sum of all horizontal forces should be zero):
$\begin{gathered}
   \Rightarrow F\cos \theta - f = 0 \\
   \Rightarrow F\cos \theta = f = 0.2R \\
   \Rightarrow F\cos \theta = 0.2(2000 - F\sin \theta ){\text{ (from eq}}{\text{.(i))}} \\
   \Rightarrow F\cos \theta = 400 - 0.2F\sin \theta \\
   \Rightarrow F\cos \theta + 0.2F\sin \theta = 400 \\
\end{gathered} $
Multiplying both sides by 5 then,
$\begin{gathered}
   \Rightarrow F(5\cos \theta + \sin \theta ) = 2000 \\
   \Rightarrow F = \dfrac{{2000}}{{5\cos \theta + \sin \theta }}{\text{ - - - - - - - - - - - - - - - - - eq(ii)}} \\
\end{gathered} $
Work done by the person pulling the box is:
= component of force acting in the direction of displacement x total displacement of the box
$\begin{gathered}
   = F\cos \theta \times d \\
   = \left( {\dfrac{{2000}}{{5\cos \theta + \sin \theta }}} \right) \times \cos \theta \times 20{\text{ (from eq(ii))}} \\
   = \dfrac{{40000\cos \theta }}{{5\cos \theta + \sin \theta }} \\
  {\text{dividing both numerator and denominator by cos}}\theta \\
   = \dfrac{{8000}}{{1 + 0.2\tan \theta }}{\text{ - - - - - - - - - - - - - - - - - - - - - eq}}{\text{.(iii)}} \\
\end{gathered} $
For minimizing the Force:
$\begin{gathered}
  \dfrac{{dF}}{{d\theta }} = 0 \\
   \Rightarrow \dfrac{{d\left( {\dfrac{{2000}}{{5\cos \theta + \sin \theta }}} \right)}}{{d\theta }} = 0 \\
   \Rightarrow 2000 \times \dfrac{{d\left( {\dfrac{1}{{5\cos \theta + \sin \theta }}} \right)}}{{d\theta }} \times \dfrac{{d\left( {5\cos \theta + \sin \theta } \right)}}{{d\theta }}{\text{ (by chain rule)}} \\
   \Rightarrow {\text{2000}} \times \dfrac{{ - 1}}{{{{\left( {5\cos \theta + \sin \theta } \right)}^2}}} \times \left( { - 5\sin \theta + \cos \theta } \right) = 0 \\
   \Rightarrow - 5\sin \theta + \cos \theta = 0 \\
   \Rightarrow \tan \theta = \dfrac{1}{5} = 0.2 \\
\end{gathered} $
So, by putting this value of $\tan \theta $in equation (iii), work done expression.

$\begin{gathered}
  W = \dfrac{{8000}}{{1 + 0.2 \times 0.2}} = \dfrac{{8000}}{{1 + \dfrac{1}{{25}}}} = \dfrac{{8000 \times 25}}{{26}} \\
  W = 7692J \\
\end{gathered} $

Note:
Mathematically, the expression for work done is given by:
 Where,
W is the amount of work done by the force.
F is the magnitude of force and d is the displacement caused by the force
θ is the angle between the force vector and the displacement vector
Important points:
The formula for work done is given by a dot product hence it is a scalar quantity.
The SI unit of work is Joule (J).