91Ó°ÊÓ

Weight,${}^{\mathbf{W}\mathbf{=}\mathbf{45}\mathbf{}\mathbf{N}}$

Coefficient of static friction,${}^{{\mathbf{μ}}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}}$

Coefficient of kinetic friction,${}^{{\mathbf{μ}}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{34}}$

Inclined angle,${}^{\mathbf{θ}\mathbf{=}{\mathbf{15}}^{\mathbf{0}}}$

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. 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:

$F_{net}=∑ma$

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

Free body Diagram of sled over the inclined plane:

The maximum static frictional force, $f_{max}={\mathrm{μ}}_{s}N,$

By using Newton’s 2nd law of motion along vertical direction,

$N=W\cos \left({15}^0\right)$

Thus,${}^{f_{max}={\mathrm{μ}}_{s}N={\mathrm{μ}}_s(W\cos \left(15\right))=(0.50)\left(45\right)\cos \left({15}^0\right)=21.73N}$

Similarly, the maximum kinetic friction,

$$\begin{gathered} f_{max}={\mathrm{μ}}_{k}N \\ ={\mathrm{μ}}_k(W\cos \left(15\right)) \\ =(0.34)\left(45\right)\cos \left(15\right) \\ =14.78N \end{gathered}$$

If force P is applied along the inclination, then it’s effect on the frictional force by using Newton’s 2nd law along horizontal direction (x) is,

$∑F_x=m{a}_x$

since, the block is not moving, ${}^{a_x=0}$

$$\begin{gathered} f_s-P-W\sin \left(\mathrm{θ}\right)=0 \\ {f}_s=P+W\sin \left(\mathrm{θ}\right) \end{gathered}$$

Since, ${}^{f_s}$is along the opposite direction of it should be ${}^{16.6\hat{\mathrm{i}}}$. This force is less than maximum static frictional force. So, the static frictional force is equivalent to the ${}^{16.6N}$force along $\hat{\mathrm{i}}$

Similarly, here also ${}^{f_s<f_{max}}$So, the static frictional force is equivalent to the 16.6N force alongrole="math" localid="1654156238627" $\hat{i}$.

Since, ${}^{f_s>f_{max}}$, then the block begins to slide. Therefore, the kinetic frictional comes to play. So, the frictional force or kinetic frictional force acting on the block is role="math" localid="1654156555096" ${}^{14.78\mathrm{N}}$along the