Atwood machine with friction on one block

[natalydj83]

Please help with friction and acceleration

hw Problem:
A block of mass 1 kg rests on a tabletop with coeficient of kinetic frictionequal to 0.3. The block is conected by a string which passes over a frictionless pulley to a second block of mass 2kg which hangs vertically from the string. The acceleration of the two block is...?

mass(1)=1kg
mass(2)=2kg
uk=0.3(coeficient)
Fr(friction)=uk*Fn(normal force)

My solution:
(x)F=ma

box 1
component (x)=Ft(force tension)-Ffr(friction)=ma
component (y)=Fn(normal force)-mg(weight)=m(a=0)
Fn=mg
Fn=1kg*9.8=9.8N

component (x)=Ft-uk*Fn=ma
component (x)=Ft=ma-uk*Fn
Ft=ma-0.3*9.8=ma-2.94
Ft1=ma-2.94N

Box 2
component (y)=mg-Ft=ma
mg-ma=Ft

Now: Ft1=Ft2
m(1)a-2.94=m(2)g-m(2)a
a=(m(2)g+2.94)/m1+m2

a=(2*(9.8)+2.94)/2+1=7.51m/s^2

IS THIS SOLUTION RIGHT? If not please indicate where i made a mistake

 

[wais]

and how to fix it

Your solution is correct. The only thing I would suggest is to label the directions of the forces and acceleration to make it clearer. Also, it's always a good idea to check your units to make sure they are consistent throughout the problem. In this case, the units for the tension force are in Newtons (N) and the units for the acceleration are in meters per second squared (m/s^2). Overall, you did a great job breaking down the problem and using the correct equations to solve for the acceleration.