91Ó°ÊÓ

Mass of block 1,$m_1=2.0\mathrm{kg}$.

Mass of block 2,$m_2=1.0\mathrm{kg}$.

Force applied on block 2,F=20 N .

Angle at which force is applied,$\mathrm{θ}=30.0°$.

The problem deals with the Newton’s laws of motion which describe the relations between the forces acting on a body and the motion of the body.Applying Newton's laws, we can solve the given problem.

For the$m_2==1.0\mathrm{kg}$block, application of Newton's laws results in

$$\begin{gathered} F\cos \mathrm{θ}-T-f_k=m_{2}axaxis \\ {F}_N-F\sin \mathrm{θ}-m_{2}g=0yaxis \end{gathered}$$

Since $f_k={\mathrm{μ}}_k{F}_N$, these equations can be combined into an equation to solve for:

$F\left(\cos \mathrm{θ}-{\mathrm{μ}}_k\sin \mathrm{θ}\right)-T-{\mathrm{μ}}_k{m}_{2}g=m_{2}a$

Similarly (but without the applied push) we analyze the$m_1=2.0kg$block:

$$\begin{gathered} T-{f^{\text{'}}}_k=m_{1}axaxis \\ {F^{\text{'}}}_N=m_{1}g=0yaxis \end{gathered}$$

Using $f_k={\mathrm{μ}}_k{F^{\text{'}}}_N$, the equations can be combined:

$T-{\mathrm{μ}}_k{m}_{1}g=m_{1}a$

Subtracting the two equations for and solving for the tension, we obtain

$$\begin{gathered} T=\frac{m_1\left(\cos \mathrm{θ}-{\mathrm{μ}}_k\sin \mathrm{θ}\right)}{m_1+m_2}F \\ ⇒T=\frac{\left(2.0\mathrm{kg}\right)\left[\cos 35°-\left(0.20\right)\sin 35°\right]}{2.0\mathrm{kg}+1.0\mathrm{kg}}\left(20\mathrm{N}\right) \\ ⇒T=9.4N \end{gathered}$$