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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A block of mass m = 2kg is raised vertically upwards by means of a massless string through a distance of S=4m with a constant acceleration a=2.2ms$^{-2}$.Find the net work done on the block.

$\begin{array}{1 1}17.6J\\16.6J\\18.6J\\19.6J\end{array} $

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Answer : 17.6J
From the free body diagram
$T-mg =ma$
$\Rightarrow T=m(a+g)$
$\quad\quad=2\times (2.2+9.8)$
$\therefore$ Work done by tension is (T and S are in the same direction)
$W_1=TS\cos 0^{\large\circ}$
$\Rightarrow 24\times 4\times 1=96J$
Since the gravitional force $mg$ and displacement S are in opposite directions,work done by gravity is
$W_2=mgS\cos 180^{\large\circ}$
$\Rightarrow -2\times 9.8\times 4 =-78.4J$
Net work done $W=W_1+W_2=96-78.4=17.6J$
answered Aug 28, 2014 by sreemathi.v

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