A cord is used to lower vertically a block of mass $M$ for a distance $d$ at a constant downward acceleration $g / 6$ . Then the work done by the cord on the block is

\frac{5 M g d}{6}

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\frac{- 5 M g d}{6}

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2 M g d

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- 2 M g d

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Solution

The correct option is B $\frac{- 5 M g d}{6}$

Applying Newton's $2^{n d}$ law,
$M g - T = M a \Rightarrow M g - T = \frac{M g}{6}$
$\Rightarrow T = \frac{5 M g}{6}$

Work done by the cord will be,
$W = - T d = - \frac{5 M g d}{6}$

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A cord is used to lower vertically a block of mass M for a distance d at a constant downward acceleration g/6. Then the work done by the cord on the block is

The correct option is B −5Mgd6 Applying Newton's 2nd law, Mg−T=Ma⇒Mg−T=Mg6 ⇒T=5Mg6 Work done by the cord will be, W=−Td=−5Mgd6

A cord is used to lower vertically a block of mass $M$ for a distance $d$ at a constant downward acceleration $g / 6$ . Then the work done by the cord on the block is

The correct option is B $\frac{- 5 M g d}{6}$

Applying Newton's $2^{n d}$ law,
$M g - T = M a \Rightarrow M g - T = \frac{M g}{6}$
$\Rightarrow T = \frac{5 M g}{6}$

Work done by the cord will be,
$W = - T d = - \frac{5 M g d}{6}$