An object of 1 kg is slowly lifted to 100 m above the ground. Calculate the work done by gravity on the object.
Take acceleration due to gravity $= 10 m s^{- 2} .$

100 J

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- 100 J

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1000 J

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- 1000 J

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Solution

The correct option is D - 1000 J
Let the upward direction be positive
Force applied by the gravity is equal to the weight of the object, as the object is lifted slowly.
F = W = -mg
Minus sign shows that force is downwards
Displacement , s = 100 m
W = F s
W = - mgs
$W = 10 \times 10 \times 100 = - 1000 J$
Since force and displacement are in the opposite direction, work done is negative.

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