91Ó°ÊÓ

The mass of the bucket is: ${\mathrm{m}}_{\mathrm{b}}=65\mathrm{kg}$

The mass of the box is: $\mathrm{m}=80\mathrm{kg}$

The mass of the gravel bag is: ${\mathrm{m}}_{\mathrm{b}}=50\mathrm{kg}$

The coefficient of static friction on the box is: ${\mathrm{μ}}_{\mathrm{s}}=0.700$

The coefficient of kinetic friction on the box is: ${\mathrm{μ}}_{\mathrm{k}}=0.400$

The distance descended by the bucket is:$\mathrm{h}=2\mathrm{m}$

The friction force varies with the coefficient of friction between object and surface. For a smooth surface, it is less and for rough surfaces, it is very high.

(a)

There is no movement between the gravel bag and box so the friction force on the bag of gravel will be zero.

The friction force on the box is given as:

${\mathrm{f}}_{\mathrm{b}}={\mathrm{m}}_{\mathrm{b}}\mathrm{g}$

${\mathrm{f}}_{\mathrm{b}}$is the friction force on the box, ${\mathrm{m}}_{\mathrm{b}}$is the mass of the bucket.

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{f}}_{\mathrm{b}}=\left(65\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ {\mathrm{f}}_{\mathrm{b}}=637\mathrm{N} \end{gathered}$$

Therefore, the friction force for gravel of bag and box are 0 N and 637 N.

(b)

Apply the energy conservation to find the speed of the bucket.

${\mathrm{m}}_{\mathrm{b}}\mathrm{gh}-{\mathrm{μ}}_{\mathrm{k}}\left(\mathrm{m}+{\mathrm{m}}_{\mathrm{g}}\right)\mathrm{gh}=\frac{1}{2}{\mathrm{m}}_{\mathrm{b}}{\mathrm{v}}^2$

Substitute all the values in the above equation.

$$\begin{gathered} \left(65\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2\mathrm{m}\right)-\left(0.4\right)\left(80\mathrm{kg}+50\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2\mathrm{m}\right)=\frac{1}{2}\left(65\mathrm{kg}\right){\mathrm{v}}^2 \\ \mathrm{v}=2.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the bucket is $2.8\mathrm{m}/\mathrm{s}$.