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A box is dragged across a floor by a rope which makes an angle of ${45^ \circ }$ with the horizontal. The tension in the rope is $100\,N$ while the box is dragged $10\,m$ . The work done is:
(A) $607.1\,J$
(B) $707.1\,J$
(C) $1414.2\,J$
(D) $900\,J$

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Last updated date: 20th Jun 2024
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Answer
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Hint To solve the question, you first need to visualize the situation and have a look at all the forces acting on the box. Since the rope makes some angle with the horizontal, only a component of the tension of rope is responsible for the displacement of the box. Find that component and then, the work done by this component of the tension of rope, which will be the answer to the question.

Complete step by step answer
We will proceed with the solution exactly as explained in the hint section of the solution to the question. Let us first draw the FBD (Free Body Diagram) of the box and observe all the forces acting on the box as well.

In the diagram, W is the weight of the box
N is the normal reaction force due to the common contact of surface
$T$ is the tension of the rope which is pulling the box
And ${T_h}$ is the horizontal component of the tension and ${T_v}$ is the vertical component of the tension
If we have a look at the FBD of the box, we can clearly see that there is only one force acting on the box horizontally, and that is the horizontal component of the tension of rope, ${T_h}$
The question has already told us that the value of the tension $T$ is:
$T = 100\,N$
The angle of the rope with the ground is given to be ${45^ \circ }$
So, the horizontal component of the tension comes out to be:
$
  {T_h} = T\cos {45^ \circ } \\
   \Rightarrow {T_h} = 100 \times \dfrac{1}{{\sqrt 2 }} \\
   \therefore{T_h} = 70.71\,N \\
 $
Now, the question has told us that the box has been dragged by a distance of $10\,m$
We know that work is nothing but the product of the distance travelled by an object and the force acting in the direction of motion, so, we can write it as:
$WD = {T_h}s$
Where $WD$ is work done,
${T_h}$ is the horizontal component of the tension in the rope
$s$ is the distance travelled by the block in the direction of the force
Substituting the values, we get:
$
  WD = 70.71 \times 10 \\
  WD = 707.1\,J \\
 $

Hence, we can see that the correct option is option (B) since the value matches with what we found out by solving the question.

Note Many students do not consider the components of the force and directly multiply the tension in the rope with the distance travelled by the box and reach at a wrong answer. The right approach is to only consider the force which is acting in the direction of the motion or against the direction of the motion to find the right answer.