91Ó°ÊÓ

Mass of block: 1.5 kg

Horizontal force applied:$F=\left(1.8t\right)\stackrel{Áåœ}{i}N$

The acceleration of the box

a=0 For $0≤t≤2.8s$and $a=\left(1.2t-2.4\right)\stackrel{Áåœ}{i}m/s^2$for t>2.8 s

The problem deals with the Newton’s second law of motion which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object. Use the Newton's second law to the horizontal axis (positive in the direction of motion)

Formula:

F=ma

(a)

The box doesn't move until time is 2.8 s, which is when the applied force $\stackrel{→}{F}$reaches a magnitude of $F=\left(1.8\right)\left(2.8\right)=5.0\mathrm{N}$, implying therefore that $f_{s,max}=5.0\mathrm{N}$. Analysis of the vertical forces on the block leads to the observation that the normal force magnitude equals the weight

$$\begin{gathered} F_N=mg \\ =15\mathrm{N} \end{gathered}$$.

Thus,

$$\begin{gathered} {\mathrm{μ}}_s=f_{s,m}I{F}_N \\ =0.34 \end{gathered}$$

(b)

$$\begin{gathered} F-f_k=ma \\ ⇒1.8t-f_k=\left(1.5\right)\left(1.2t-2.4\right) \end{gathered}$$

Thus, we find$f_k=3.6\mathrm{N}$.

Therefore,${\mathrm{μ}}_k=f_k/F_N=0.24$