The work done against force of friction is: (consider ideal quantites )

8.7 J

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10.7 J

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7.8 J

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12.7 J

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Solution

The correct option is A $8.7 J$
Work against friction is
$W_{f} = f d = μ N d$

$= μ m g cos θ d (∵ N = m g cos θ)$

$= 0.1 \times 1 k g \times 10 m s^{- 2} \times cos 30^{\circ} \times 10 m$

$= 8.7 J$ .

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Calculate the work done in pushing the card through distance of 15 against the force of friction equal to 250 Newton state the type of work done

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