91Ó°ÊÓ

From the graph,

$$\begin{gathered} {\mathrm{μ}}_{k_1}=0at{a}_1=3.0\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{μ}}_{{\mathrm{k}}_1}=0.20\mathrm{at}{\mathrm{a}}_2=0\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{μ}}_{{\mathrm{k}}_1}=0.40\mathrm{at}{\mathrm{a}}_3=-3.0\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

To find the angle ${}^{\mathbf{θ}}$ use Newton's 2nd law of motion. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\underset{}{\overset{}{∑ma}}$

where, F is the net force, m is mass and a is an acceleration.

Free Body Diagram of the Block:

By using the Newton’s 2nd law along the vertical direction to the block A,(positive x axis along the right and the positive y along the vertical direction),

$$\begin{gathered} F_N-F\sin \mathrm{θ}-mg=0 \\ {F}_N=F\sin \mathrm{θ}+mg \end{gathered}$$

Thus, the kinetic frictional force,

$$\begin{gathered} f_k={\mathrm{μ}}_k{F}_N \\ ={\mathrm{μ}}_k\left(F\sin \mathrm{θ}+mg\right) \end{gathered}$$

Similarly, to the horizontal direct

$$\begin{gathered} F\cos \mathrm{θ}-f_k=ma \\ F\cos \mathrm{θ}-{\mathrm{μ}}_k\left(F\sin \mathrm{θ}+mg\right)=ma \\ a\frac{F}{m}\left(\cos \mathrm{θ}-{\mathrm{μ}}_k\sin \mathrm{θ}\right)-{\mathrm{μ}}_{k}g \\ \mathrm{At},{\mathrm{μ}}_k=0and{a}_1=3.0\mathrm{m}/{\mathrm{s}}^2, \\ 3.0=\frac{F}{m}\left(cos\mathrm{θ}\right) \\ \mathrm{At},{\mathrm{μ}}_{k_2}=0.20and{a}_2=0\mathrm{m}/{\mathrm{s}}^2 \\ 0=\frac{F}{m}\left(cos\mathrm{θ}-\left(0.20\right)\sin \mathrm{θ}\right)-\left(0.20\right)g \\ 0=\frac{F}{m}\left(\cos \mathrm{θ}\right)-\left(020\right)\frac{F}{m}\sin \mathrm{θ}-\left(0.20\right)g \\ 0=3.0-\left(0.20\right)\frac{F}{m}\sin \mathrm{θ}-\left(0.20\right)g \end{gathered}$$ (i)

From equation (i),

$$\begin{gathered} \left(0.20\right)\frac{F}{m}\sin \mathrm{θ}=3.0-\left(0.20\right)g \\ \left(0.20\right)\frac{\left(3.0\right)}{\left(\cos \mathrm{θ}\right)}\sin \mathrm{θ}=3.0-\left(0.20\right)g \\ \left(0.6\right)\tan \mathrm{θ}=1.04 \\ \tan \mathrm{θ}=\frac{1.04}{0.6} \\ =1.73 \\ \mathrm{θ}=60° \end{gathered}$$

Hence, the magnitude of an angle is$60°$