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**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:

\[\sum {{F_x} = 0,\sum {{F_Y} = 0} } \]

Then, find the work done by the expression,

Work done = component of force acting in the direction of displacement $ \times $ total displacement of the box

**Complete step by step answer:**

Let’s start with a free body diagram.

Given that,

Weight of the box = 2000 N

Friction coefficient $ = \mu = 0.2$

The box slowly slid through the floor about 20 m.

From free body diagram the forces acting on the boxes are:

Weight of the box is acting downwards,

W = mg = 2000 N

Normal reaction force will be acting upwards.

The force exerted by a person will have two components given by \[Fcos\theta \] and \[Fsin\theta \].

The frictional force acting on the box $ = {f_r} = \mu N$

To make the box to remain in its equilibrium condition, the sum of all forces acting on the box is zero.

Thus, on applying equilibrium conditions.

$\sum {{F_y} = 0} $ (forces acting in vertical direction)

$ \Rightarrow N + F\sin \theta - 2000 = 0$

$ \Rightarrow N = 2000 - F\sin \theta $

Similarly,

$\sum {{F_x} = 0} $

$ \Rightarrow F\cos \theta - {f_r} = 0$

$ \Rightarrow F\cos \theta = {f_r}$

$ \Rightarrow F\cos \theta = \mu N = 0.2N$

Substituting, the value of normal force N, we get

$ \Rightarrow F\cos \theta = 0.2(2000 - F\sin \theta )$

$ \Rightarrow F\cos \theta + 0.2F\sin \theta = 400$

$ \Rightarrow F = \dfrac{{400}}{{\cos \theta + 0.2\sin \theta }}$

Therefore, work done in pushing the block is given by

$ \Rightarrow W = F.s$

$ \Rightarrow W = F\cos \theta \times s$

$ \Rightarrow W = 20\left( {\dfrac{{400}}{{\cos \theta + 0.2\sin \theta }}} \right)\cos \theta $

$ \Rightarrow W = \dfrac{{8000}}{{1 + 0.2\tan \theta }}joule$

**Note:**

The formula for work done is given by a dot product, hence it is a scalar quantity. Mathematically it is given by $W = \vec F.\vec s$

The SI unit of work is joule(J).

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