What Is the Acceleration of Two Blocks in a Frictional System?

[natalydj83]

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

 

[Ba]

On box 1 you made a mistake, Ft-uk*Fn=ma then you went to Ft=ma-uk*Fn. It should be Ft=ma+uk*Fn.

 

[drnikitin]

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

The net force is F=2*9.8+0.3*9.8=22.54 N. The net mass is m=2+1=3kg. So the acceleration is a=F/m = 22.54/3=7.5 m/s^2. Thus your answer is correct