- #1
natalydj83
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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
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