Respuesta :
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.
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.