Question

A 1.5 kgbox is initially at rest on a horizontal surface when at t =0 a horizontal force $\overrightarrow{\mathbf{f}}\mathbf{=}\left(1.8t\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{N}$(with tin seconds) is applied to the box. The acceleration of the box as a function of time tis given b role="math" localid="1660971208695" $\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{for}\mathbf{}\mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{\le }\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{s}$and:$\overrightarrow{\mathbf{a}}\mathbf{=}\left(1.2t-2.4\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ for t>2.8 s(a) what is the coefficient of static friction between the box and the surface? (b) What is the coefficient of kinetic friction between the box and the surface?

Short answer

a) Coefficient of static friction between the box and surface is

b) Coefficient of kinetic friction between the box and the surface is 0.24

Step-by-step solution

Given

Mass of block: 1.5 kg

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

The acceleration of the box

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

Understanding the concept

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

Calculate the coefficient of static friction between the box and the surface

(a)

The box doesn't move until time is 2.8 s, which is when the applied force $\overrightarrow{F}$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,

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

Calculate the coefficient of kinetic friction between the box and the surface

(b)

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

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

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