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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

Sagot :

LammettHash

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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