91Ó°ÊÓ

Mass of the block, $\mathrm{m}=68\mathrm{kg}$

Coefficient of static friction,${\mathrm{μ}}_{\mathrm{a}}=0.50$

Coefficient of kinetic friction, ${\mathrm{μ}}_{\mathrm{k}}=0.35$

Angle,$\mathrm{θ}=15°$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion along vertical and horizontal direction.

Formula:

${\mathrm{F}}_{\mathrm{net}}=\underset{}{∑}\mathrm{ma}$

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

Free body diagram of the block:

By using Newton’s 2nd law along y direction, block is not moving along y.So, the acceleration is 0, it gives,

$$\begin{gathered} \underset{}{∑}{\mathrm{F}}_{\mathrm{x}}=0 \\ \mathrm{N}-\mathrm{mg}+\mathrm{F}\sin \left(15\right)=0 \\ \mathrm{N}=\mathrm{mg}-\mathrm{F}\sin \left(15\right) \\ \mathrm{Similarly},\mathrm{along}\mathrm{x}\mathrm{direction}, \\ \underset{}{∑}{\mathrm{F}}_{\mathrm{x}}=0 \\ \mathrm{since}\mathrm{block}\mathrm{is}\mathrm{at}\mathrm{rest},\mathrm{a}=0 \\ \mathrm{Fcos}\left(15\right)-{\mathrm{f}}_{\mathrm{s}}=0 \\ {\mathrm{f}}_{\mathrm{s}}=\mathrm{Fcos}\left(15\right) \\ \mathrm{The}\mathrm{frictional}\mathrm{force}\mathrm{is}, \\ {\mathrm{f}}_{\mathrm{s}}={\mathrm{μ}}_{\mathrm{s}}\mathrm{N}={\mathrm{μ}}_{\mathrm{s}}(\mathrm{mg}-\mathrm{Fsin}(15\left)\right) \\ \mathrm{Equating}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right), \\ \mathrm{Fcos}\left(15\right)={\mathrm{μ}}_{\mathrm{s}}(\mathrm{mg}-\mathrm{Fsin}(15\left)\right) \\ \mathrm{Fcos}\left(15\right)+{\mathrm{μ}}_{\mathrm{s}}\mathrm{Fsin}\left(15\right)={\mathrm{μ}}_{\mathrm{s}}\mathrm{mg} \\ \mathrm{F}=\frac{{\mathrm{μ}}_{\mathrm{s}\mathrm{mg}}}{\cos \left(15\right)+{\mathrm{μ}}_{\mathrm{s}}\sin \left(15\right)} \\ =\frac{\left(0.50\right)\left(68\right)\left(9.81\right)}{\cos \left(15\right)+\left(0.50\right)\sin \left(15\right)} \\ =304\mathrm{N} \end{gathered}$$

Therefore, the minimum force magnitude is required from the rope to start the crate moving is 304N.

By using Newton’s 2nd law along y direction,block is not moving along y.So, the acceleration is 0, it gives,

$$\begin{gathered} \underset{}{∑{\mathrm{F}}_{\mathrm{y}}=0} \\ \mathrm{N}=\mathrm{mg}+\mathrm{F}\sin \left(15\right)+0 \\ \mathrm{N}=\mathrm{mg}-\mathrm{F} \end{gathered}$$

The kinetic frictional force is,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{k}}={\mathrm{μ}}_{\mathrm{k}}\mathrm{N} \\ ={\mathrm{μ}}_{\mathrm{k}}\left(\mathrm{mg}-\mathrm{Fsin}\left(15\right)\right) \end{gathered}$$

According to the Newton’s 2nd law along horizontal direction i.e. x is as the block movingalong x.Let, the acceleration of the block isa,

$$\begin{gathered} \underset{}{∑{\mathrm{F}}_{\mathrm{x}}={\mathrm{ma}}_{\mathrm{x}}} \\ \mathrm{Fcos}\left(15\right)-{\mathrm{f}}_{\mathrm{x}}=\mathrm{ma} \\ \mathrm{a}=\frac{\mathrm{Fcos}\left(15\right)-{\mathrm{μ}}_{\mathrm{x}}\left(\mathrm{mg}-\mathrm{Fsin}\left(15\right)\right)}{\mathrm{m}} \\ =\frac{\left(304\right)\cos \left(15\right)-{\mathrm{μ}}_{\mathrm{k}}\left(\mathrm{mg}-\mathrm{Fsin}\left(15\right)\right)}{\mathrm{m}} \\ =1.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the magnitude of the initial acceleration of the crate is 1.3 m/s2.