91Ó°ÊÓ

Weight of the crate:$W=mg=165\mathrm{N}$

Coefficient of static friction: ${\mathrm{μ}}_s=0.51$

Coefficient of kinetic friction:${\mathrm{μ}}_k=0.32$

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. Also it involves friction force. Draw a free body diagram for the given situation and then apply Newton’s first and second law.

Formula:

Frictional force is given by,

${\mathbf{f}}_{\mathbf{k}}\mathbf{=}{\mathbf{μ}}_{\mathbf{k}}{\mathbf{F}}_{\mathbf{N}}$

(a)

The push or the $F_{applied}$to get the crate just moving the applied force must be as big as the maximum force of static friction = $f_{s.max}={\mathrm{μ}}_s{F}_N$and here $F_N=W=165\mathrm{N}$

Hence, required push,

$$\begin{gathered} F_{applied}=f_{s.max} \\ ={\mathrm{μ}}_s{F}_N \\ ⇒F_{applied}=0.51×165\mathrm{N} \\ ⇒{\mathrm{F}}_{\mathrm{applied}}=84.15\mathrm{N} \end{gathered}$$

(b)

While in motion, constant velocity (zero acceleration) is maintained if the push is equal to the kinetic friction force

$$\begin{gathered} F\left(applied\right)=f_k \\ ={\mathrm{μ}}_s{F}_N \\ ⇒F_{\mathit{a}\mathit{p}\mathit{p}\mathit{l}\mathit{i}\mathit{e}\mathit{d}}=0.32×165\mathrm{N} \\ ⇒F_{\mathit{a}\mathit{p}\mathit{p}\mathit{l}\mathit{i}\mathit{e}\mathit{d}}\mathit{=}52.8\mathrm{N} \end{gathered}$$

(c)

Mass of the crate,

$$\begin{gathered} m=\frac{165\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =16.84\mathrm{kg} \end{gathered}$$

The acceleration, using the push from part (a), is:

$a=\frac{84.15\mathrm{N}-52.8\mathrm{N}}{16.84\mathrm{kg}}$

$⇒a=1.87m/s^2$