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# A force F acting on an object varies with distance x as shown in figure. The force is in N and x in m. The work done by the force in moving the object from $x = 0{\text{ to }}x = 6m$is:(A) $13.5{\text{ J}}$(B) $10{\text{ J}}$(C) $15{\text{ J}}$(D) $20{\text{ J}}$

Last updated date: 02nd Aug 2024
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Hint: Work is said to be done when a force produces motion in an object. The work done can be defined as the product of the force F and the displacement produced in the object, x. In a graphical form, F and x can be multiplied, and the area gives us the total value of the work done.

In a graphical representation of F vs x, we can integrate the curve and/or find the area enclosed between the two quantities.
This is the equivalent of multiplying in both quantities.
So, In the given graph we can divide the complex shape formed, into 2 simple shapes namely: A rectangle and a triangle.
Finding out areas of both these figures and adding them gives the value of total work done.
Area of triangle, ${A_1} = \dfrac{1}{2}bh$
From the graph, $b = 6 - 3 = 3$
$h = 3$
$\therefore {A_1} = \dfrac{1}{2} \times 3 \times 3$
${A_1} = 4.5Nm$
Now, Area of rectangle (square here)
${A_2} = {a^2}$
Where the side length, $a = 3$
${A_2} = {3^2}$
${A_2} = 9Nm$
The total work done, W is defined as:
$W = {A_1} + {A_2}$
$W = 4.5 + 9$
$W = 13.5Nm{\text{ or }}13.5J$

The correct option is (A).