91影视

Angle = 35⁰

Acceleration of the child = 0.50 m/s²

The frictional force is given by the product of the coefficient of friction and the normal reaction. Newton鈥檚 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object鈥檚 mass. Using these concepts, the problem can be solved.

In case of 鈥渄ownhill鈥� (positive), then Newton鈥檚 law gives

$mgsin胃-f_k=ma$

for the sliding child,

$$\begin{gathered} f_k={渭}_k{F}_N={渭}_{k}mg \\ a=g\left(\text{sin}胃-{渭}_k\text{cos}胃\right) \\ =-0.50{\text{m/s}}^{\text{2}} \\ g\left(\text{sin}胃-{渭}_k\text{cos}胃\right)=-0.50{\text{m/s}}^{\text{2}} \end{gathered}$$

Rearranging for the${渭}_k$, we get

${渭}_k=\frac{1}{\text{cos}胃}\left(\frac{0.50{\text{m/s}}^{\text{2}}}{g}+\text{sin}胃\right)$

Substituting the value of${渭}_k$

Withlocalid="1661258346840" $胃={35}^0$, we solve for the coefficient .

$$\begin{gathered} {渭}_k=\frac{1}{\text{cos}35}\left(\frac{0.50{\text{m/s}}^{\text{2}}}{9.8}+\text{sin}35\right) \\ =0.76 \end{gathered}$$

Therefore, the coefficient of friction between the child and slope is 0.76.