A 0.200 kg object slides on a horizontal surface where the coefficient of kinetic friction is 0.0800. Taking g as 9.80 m/s squared, what is the frictional force on it?

1. On a horizontal surface the normal force equals the weight, so N = m g. 2. Kinetic friction is f = mu_k m g. 3. Substitute: f = (0.0800)(0.200 kg)(9.80 m/s squared). 4. First take (0.200 kg)(9.80 m/s squared) = 1.96 N, the weight of the object. 5. Then f = (0.0800)(1.96 N) = 0.157 N. 6. Quoting 1.96 N gives the weight and forgets the coefficient altogether. 7. Quoting 0.0160 N multiplies the mass by the coefficient but leaves out g. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section 16.7 Damped Harmonic Motion_