g A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 3.00 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do

The desk is in equilbrium, so Newton's second law gives

∑ F (horizontal) = p - f = 0

∑ F (vertical) = n - mg = 0

==>   n = mg

==>   p = f = µn = µmg = 0.400 (80.0 kg) g = 313.6 N

The student pushes the desk 3.00 m, so she performs

W = (313.6 N) (3.00 m) = 940.8 Nm ≈ 941 J

of work.

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