91Ó°ÊÓ

The given data can be listed below as,

• The mass of the crate is, $m=45.0kg$.
• The force when crate just begin to move is, ${\mathrm{F}}_{\mathrm{s}}=313\mathrm{N}$.
• The force to keep crate moving is, ${\mathrm{F}}_{\mathrm{k}}=208\mathrm{N}$.
• The steady speed of crate is, v = 25.0 m/s .

Friction is an opposing force which resist one surface to slide over another surface when two surface are in contact with each other.

The expression for the maximum static force can be expressed as,

$$\begin{gathered} F_s={\mathrm{μ}}_{s}mg \\ {\mathrm{μ}}_s=\frac{F_s}{mg} \end{gathered}$$

Here ${\mathrm{μ}}_s$ is the coefficient of static friction, $F_s$ is the static force, m is the mass of the crate, and is acceleration due to gravity.

Substitute, 313 N for , 45.0 kg for m, $9.8m/s^2$for g in the above equation.

$$\begin{gathered} {\mathrm{μ}}_{\mathrm{s}}=\frac{313\mathrm{N}}{45.0\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2} \\ =0.700 \end{gathered}$$

Hence, required coefficient of static friction is, 0.700.

The expression for the maximum kinetic force can be expressed as,

$$\begin{gathered} F_k={\mathrm{μ}}_{k}mg \\ {\mathrm{μ}}_k=\frac{F_k}{mg} \end{gathered}$$

Here ${\mathrm{μ}}_k$is the coefficient of kinetic friction, $F_k$ is the maximum kinetic force.

Substitute 208 N for $F_k$, 45.0 kg for m, $9.8m/s^2$for in the above equation.

$$\begin{gathered} {\mathrm{μ}}_{\mathrm{k}}=\frac{208\mathrm{N}}{45.0\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2} \\ =0.471 \end{gathered}$$

Hence, required coefficient of kinetic friction is, 0.471.

The expression for the pushing force on crate according to Newton’s second law can be expressed as,

$F=ma+{\mathrm{μ}}_{k}mg$

Here m is the mass, a is the acceleration, and ${\mathrm{μ}}_k$ is the coefficient of kinetic friction, and g is the acceleration due to gravity.

Substitute 45.0 kg for m , 0.471 for${\mathrm{μ}}_k$, and $1.10\mathrm{m}/{\mathrm{s}}^2$ for a , 9.8$\mathrm{m}/{\mathrm{s}}^2$ for g in the above equation.

$$\begin{gathered} \mathrm{F}=\left(45.0\mathrm{kg}\right)\left(1.10\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.471\right)\left(45.0\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =257.2\mathrm{N} \end{gathered}$$

Hence, required force is, 257.2 N.

Part (i)

The expression for static force on the moon can be expressed as,

$F_s={\mathrm{μ}}_{s}mg$

Here ${\mathrm{μ}}_s$ is the coefficient of static friction, m is the mass, and g is the acceleration due to gravity on moon.

Substitute 0.700 for ${\mathrm{μ}}_s$, 45.0 kg for m , and $1.62\mathrm{m}/{\mathrm{s}}^2$ for g in the above equation.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{s}}=\left(0.700\right)\left(45.0\mathrm{kg}\right)\left(1.62\mathrm{m}/{\mathrm{s}}^2\right) \\ =51.03\mathrm{N} \end{gathered}$$

Hence, required magnitude to move is, 51.03.

Part (ii)

The expression for acceleration in this condition can be expressed as,

$$\begin{gathered} F-{\mathrm{μ}}_{k}mg=ma \\ a=\frac{F-{\mathrm{μ}}_{k}mg}{m} \end{gathered}$$

Here F is pushing force, m is the mass, g is the acceleration due to gravity, and ${\mathrm{μ}}_k$ is the coefficient of kinetic friction.

Substitute 257.2 N for F , 0.471 for ${\mathrm{μ}}_k$, 45.0 kg for m , $1.62\mathrm{m}/{\mathrm{s}}^2$ for g in the above equation.

$$\begin{gathered} \mathrm{a}=\frac{257.2-\left(0.471\right)\left(45.0\mathrm{kg}\right)\left(1.62\mathrm{m}/{\mathrm{s}}^2\right)}{45.0\mathrm{kg}} \\ =4.95\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, required acceleration is, $4.95\mathrm{m}/{\mathrm{s}}^2$.