A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 4.00 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? 26.7 j 1.26 kj 3.14 kj 24.0 j 128 j

1.26 kJ First, calculate the normal force. That's merely the mass of the desk multiplied by the gravitational acceleration. 80 kg * 9.8 m/s^2 = 784 kg*m/s^2 = 784 N Now calculate the force needed to slide the desk by multiplying by the coefficient of kinetic friction. So 784 N * 0.4 = 313.6 N Work is defined as force times distance, so 313.6 N * 4 m = 1254.4 Nm = 1254.4 J = 1.2544 kJ Rounding to 3 significant figures gives 1.25 kJ Of the available choices, 1.26 kJ is the closest. Looks like a mild difference in rounding. Perhaps because I rounded after all calculations were done and they may have rounded intermediate results.