Question

# A girl is carrying a school bag of $3kg$ mass on her back and moves $200m$ on a levelled road. The work done against the gravitational force will be $\left(g=10m{s}^{-2}\right)$

A

$6×{10}^{3}J$

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B

$6J$

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C

$0.6J$

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D

zero

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Solution

## The correct option is D zeroStep 1: Given dataMass of the bag (m) = $3kg$Displacement of the girl (S) = $200m$Step 2: Formula used$W=FS\mathrm{cos}\theta$, where W is the work done, F is the force acting on the body, S is its displacement and $\theta$ is the angle between the direction of force and the displacement vector.Step 3: Calculate the weight of the bagThe weight of the bag will be $=m×g$$=3×10\phantom{\rule{0ex}{0ex}}=30N$Step 4: Calculate work doneThe formula for work done is: $W=F·S\mathrm{cos}\theta$, which can be written as $W=mg·S\mathrm{cos}\theta$, since the force here will be the weight of the bag.Here, the weight acts in the vertically downward direction while displacement is in the horizontal direction. Since the displacement of the girl is in the perpendicular direction to the weight of the bag, then $\theta =90°$. Now, on substituting the given values in the formula, we get: $W=30·200·\mathrm{cos}\left(90°\right)\phantom{\rule{0ex}{0ex}}W=30·200·0\phantom{\rule{0ex}{0ex}}W=0J$Hence, option (D) is correct.

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