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A ladybug takes of with a force of 0.012 N for a distance of 6 cm. What is the work done by the ladybug during take-off?

1 Answer

Mar 7, 2018

Work = Force times distance, or $W = F \times d$ $W=Fxxd$

Explanation:

We first convert $6 c m = 0.06 m$ $6cm=0.06m$

$W = 0.012 N \times 0.06 m = 0.00072 N m = 0.00072 J$ $W=0.012Nxx0.06m=0.00072Nm=0.00072J$
(since $N m = J$ $Nm=J$ )

In scientific notation this would be:

$W = 1.2 \cdot 10^{- 2} N \times 6 \cdot 10^{- 2} m = 7.2 \cdot 10^{- 4} J$ $W=1.2*10^(-2)Nxx6*10^(-2)m=7.2*10^(-4)J$

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