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3 December, 13:34

A horizontal force pushes a 26 kg sled along a driveway for a distance of 36 m. If the coefficient of sliding friction is 0.2, what is the work done by the friction force?

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  1. 3 December, 14:37
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    Answer:1834.56 joules

    Explanation:

    Distance=36m

    Coefficient of friction=0.2

    Mass=26kg

    Acceleration due to gravity=9.8m/s^2

    Reaction=mass x acceleration due to gravity

    Reaction=26 x 9.8

    Reaction=254.8N

    Coefficient of friction=frictional force ➗ reaction

    0.2=frictional force ➗ 254.8

    Frictional force=0.2 x 254.8

    Frictional force=50.96N

    work=friction force x distance

    Work=50.96 x 36

    Work=1834.56

    Work=1834.56 joules
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