[3] A 25 kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in motion, a 127 N horizontal force keeps it moving at a constant velocity. a. Find the coefficient of static friction between the chair and the floor. b. Find the coefficient of kinetic friction between the chair and the floor.

## Step 1<br />The problem involves the concepts of static and kinetic friction. The force required to set the chair in motion is the force of static friction, while the force required to keep the chair moving at a constant velocity is the force of kinetic friction.<br /><br />## Step 2<br />The force of static friction (\(F_{s}\)) is given by the formula:<br />### \(F_{s} = \mu_{s} \times N\)<br />where \(\mu_{s}\) is the coefficient of static friction and \(N\) is the normal force. In this case, the normal force is equal to the weight of the chair, which is the product of the mass of the chair and the acceleration due to gravity (\(g\)).<br /><br />## Step 3<br />The force of kinetic friction (\(F_{k}\)) is given by the formula:<br />### \(F_{k} = \mu_{k} \times N\)<br />where \(\mu_{k}\) is the coefficient of kinetic friction.<br /><br />## Step 4<br />We can rearrange the formulas to solve for the coefficients of friction:<br />### \(\mu_{s} = \frac{F_{s}}{N}\)<br />### \(\mu_{k} = \frac{F_{k}}{N}\)<br /><br />## Step 5<br />Substituting the given values into the formulas, we get:<br />### \(\mu_{s} = \frac{165 \, \text{N}}{25 \, \text{kg} \times 9.8 \, \text{m/s}^2}\)<br />### \(\mu_{k} = \frac{127 \, \text{N}}{25 \, \text{kg} \times 9.8 \, \text{m/s}^2}\)