The relationship between the force $F$ and position $x$ of a body is as shown in figure. The work done in displacing the body from $x = 1 m$ to $x = 5 m$ will be :

30 J

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15 J

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25 J

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20 J

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Solution

The correct option is D $20 J$
$W =$ area under $F - x$ graph

From $X = 0$ to $X = 1$ , force and displacement both are positive.

Therefore work done is positive.

$W_{0} = + A r e a = + 5 J$

From $X = 1$ to $X = 3,$ force and displacement both are positive: Therefore work done is positive.
$W_{1} = + A r e a = + 15 J$
From $X = 3$ to $X = 4$ , force is negative but displacement is positive.

Therefore work done is again positive.
$∴ W_{2} = - A r e a = - 5 J$
From $X = 4$ to $X = 5$ , force and displacement both are positive.

Therefore work done is again positive.
$∴ W_{3} = + A r e a = \frac{1}{2} \times 1 \times 10 = + 5 J$
$W_{T o t a l} = W_{0} + W_{1} + W_{2} + W_{3} = 20 J$

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