Question

A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student uses a force of 15 N, what is the coefficient of kinetic friction of the floor?



dont tell me the answer just tell me step by step and let me do the rest till the end of the question

Answer 1

We're asked to find the coefficient of kinetic friction, which means that we must first find the frictional force. In order to do that, we have to use Newton's 2nd Law!\[\huge \sum \text{F}=ma\]This law only applies to one axis, whether it's y axis, x-axis, or any other conventional u-v axes (such as an incline). Applying this law horizontally, we get:\[\huge \sum \text{F}_x=ma=\text{F}_\text{applied}-\text{F}_\text{frictional}\]Perfect, but we're not exactly done yet. We know that we can always define the frictional force as\[\huge \text{F}_\text{frictional}=\mu \text{N}\]Well, now what's the normal force? By doing another summation in the y-direction, we can see that:\[\huge \sum \text{F}_y=ma=\text{N}-mg=0\]This particular equation is equal to 0 (ma = 0) because it is not accelerating in the y direction! Therefore, we can see that \(\text{N}=mg\) (Newton's 3rd Law)
Plugging this back into our previous equation, then we have:\[\huge ma=\text{F}_\text{applied}-\mu mg\]You know everything at this point except for \(\mu\)