## Because the cord has constant length, both blocks move the same number of centimeters in each second and so move with the same acceleration. To find just this acceleration, we could model the $30kg$ system as a particle under a net force. That method would not help to finding the tension,

so we treat the two blocks as separate accelerating particles.

(a) Below figure shows the free-body diagrams for the two blocks.

The tension force exerted by block $1$ on block $2$ is the same size as the tension force exerted by object $2$ on object $1$. The tension in a light string is a constant along its length, and tells how strongly the string pulls on objects at both ends.

(b) We use the free-body diagrams to apply Newtonβs second law.

For $m_{1}:βF_{x}=Tβf_{1}=m_{1}a$ or $T=m_{1}a+f_{1}$........... [1]

And also

Β Β Β $βF_{y}=n_{1}βm_{1}g=0$ or $n_{1}=m_{1}g$

Also, the definition of the coefficient of friction gives

Β Β Β $f_{1}=ΞΌn_{1}=(0.100)(12.0kg)(9.80m/s_{2})=11.8N$

For $m_{2}:βF_{x}=FβTβf_{2}=ma$...................................... [2]

Also from the $y$ component, $n_{2}βm_{2}g=0$ orΒ $n_{2}=m_{2}g$

And againΒ $f_{2}=ΞΌn_{2}=(0.100)(18.0kg)(9.80m/s_{2})=17.6N$

Substituting $T$ from equation [1] into [2], we get

Β Β Β $Fβm_{1}aβf_{1}βf_{2}=m_{2}a$ or $Fβf_{1}βf_{2}=m_{2}a+m_{1}a$

Solvind for $a$

Β Β Β $a=m_{1}+m_{2}Fβf_{1}βf_{2}β=(12.0kg+18.0kg)(68.0Nβ11.8Nβ17.6N)β=1.29m/s_{2}$

(c) From equation [1]

Β Β Β $T=m_{1}a+f_{1}=(12.0kg)(1.29m/s_{2})+11.8N=27.2N$