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

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."