The relationship between the force F and position x of a body is shown in figure. The work done in displacing the body from $x = 1$ to $x = 5$ in 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
Work done $=$ area enclosed by F-x graph.
$=$ Area of OAM + Area of ABNM+ Area of CDEN - Area of EFGH + Area of HIJ
$\frac{1}{2} \times 1 \times 10 + 1 \times 10 + 1 \times 5 - 1 \times 5 + \frac{1}{2} \times 1 \times 10$
$= 5 + 10 + 5 - 5 + 5 = 20 J$

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

The correct option is D 20 JWork done = area enclosed by F-x graph. = Area of OAM + Area of ABNM+ Area of CDEN - Area of EFGH + Area of HIJ 12×1×10+1×10+1×5−1×5+12×1×10=5+10+5−5+5=20J

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

The correct option is D 20 J
Work done $=$ area enclosed by F-x graph.
$=$ Area of OAM + Area of ABNM+ Area of CDEN - Area of EFGH + Area of HIJ
$\frac{1}{2} \times 1 \times 10 + 1 \times 10 + 1 \times 5 - 1 \times 5 + \frac{1}{2} \times 1 \times 10$
$= 5 + 10 + 5 - 5 + 5 = 20 J$