Question

A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 2.00 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?

Answer 1

628 J

Step-by-step explanation:

Force of friction x distance = work

force of friction = normal force * coeff

= ( 80 kg * 9.81 m/s^2) * .4 = 313.92 N

3139.2N * 2 m = 627.84 J = ~ 628 J