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    mikaelaq23's Avatar
    mikaelaq23 Posts: 13, Reputation: 1
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    #1

    Oct 13, 2009, 01:27 AM
    Calculating work done by friction force in physics
    A 10 kg crate is sliding at a constant speed of 7 m/s along a frictionless horizontal table, when it encounters a friction patch (s=0.8, k=0.2).

    Find the work done by friction force acting on the crate, after it has slid 5 m...

    I got 98 J, but this is wrong, not sure why, when I calculated it by
    (.2)(10kg)(9.8m/s^2)(5m)=98J... what am I doing wrong
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Oct 13, 2009, 06:09 AM

    I think you have the right appoach, except I am not familiar with the terms "s" and "k." If "k" is the coefficient of sliding friction, then your calculation is correction. However, I'm wondering if you should have used the value of "s" instead. Please review your text book and let us know whether the coefficient for sliding friction is "s" or "k."
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