hw Problem:(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Please help with hw problem finding acceleration

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