A0.350kg leather wallet rests on top of an oak table. the force required to begin to move the wallet is 2.1n

a) determine the coefficient of static friction between the wallet and table.

b) if the coefficient of kinetic friction between the wallet and table is 0.52, what is the acceleration on the wallet if the applied force pulling the wallet is 3n?

A) The normal force of the wallet is mg and the total friction is simply the normal force * the frictional force so the maximum static frictional force that can be exerted is Fnf where Fn=mg and f=frictional coefficient so the maximum amount of static friction will be .350kg*9.81m/s^2*f = 2.1N. just solve for f and you will get 2.1/(.350*9.81)=f = approximately .613=f

b) F=ma were F= F(net) the total forces acting on the system which would be the applied force - the frictional force = ma. the kinetic friction force on the system would be mgf where f = the kinetic frictional force so the equation becomes 3N-mgf=ma which gives 3N-(.350kg)(9.81m/s^2)(.52)=.350a where a is acceleration we have all the numbers for every part of the equation except a so just solve for a which gives a = (3kg*m/s^s-(.350kg*9.81m/s^2*.52))/.350kg = 3.47m/s^2 i left the units in the equation so you can see that kg*m/s^2=N so the kg*m/s^2 will cancel and you will be left with m/s^2 which is what we want for acceleration.

From negative to negative

Iknow the first part it is the average star i am not 100% sure