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Physics-

General

Easy

Question

The relation between the displacement $X$ of an object produced by the application of the variable force $F$ is represented by a graph shown in the figure. If the object undergoes a displacement from $X equals 0.5 blank m$ to $X equals 2.5 blank m$ the work done will be approximately equal to

$16 J$
$32 J$
$1.6 J$
$8 J$

The correct answer is: $16 blank J$

Work done = Area under curve and displacement axis
= Area of trapezium
$equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses s u m blank o f blank t w o blank p a r a l l e l blank l i n e s close parentheses cross times d i s t a n c e blank b e t w e e n blank t h e m$
$equals fraction numerator 1 over denominator 2 end fraction open parentheses 10 plus 4 close parentheses cross times open parentheses 2.5 minus 0.5 close parentheses equals fraction numerator 1 over denominator 2 end fraction 14 cross times 2 equals 14 blank J$
As the area actually is not trapezium so work done will be more than $14 blank J i. e.$ approximately $16 blank J$

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