91影视

The pace at which the location changes is called velocity. The average velocity is the ratio of displacement or position change (a vector quantity) to time.

From the free body diagram shows the net force acting in the vertical and horizontal direction is zero.The static friction force is equal to and opposes the conveyor force.

$f_s$is the force of static friction

${渭}_s$is the coefficient of static friction between the belt and the crate

$F$is the force applied by the conveyor

$mg$is the weight of the crate acting in downward direction

$N$is the normal reaction force

In this diagram the normal force and the weight is balanced. The force conveyor belt is equal and opposite of the static friction force.

The free body of the diagram is drawn. The net force will act on the crate in the horizontal direction as there is a relative motion between the conveyor belt and the crate.

$f_s$is the force of kinetic friction

${渭}_k$is the coefficient of kinetic friction between the belt and the crate.

$F$is the force applied by the conveyor

$mg$is the weight crate acting in downward direction

$N$is normal reaction force

Here, normal force is opposite to the direction of gravitational force.

Because the belt and the crate are in motion, kinetic friction is created between the belt and the crate. The direction of motion is reversed in this kinetic friction force. The net force acting on the crate is in the direction of right.

According to Newton's second law where $F$is force, $a$is acceleration

Mass of crate is $10kg$, coefficient of static friction between the crate and belt ${渭}_s$is $0.5$. and coefficient of kinetic friction between the crate and the belt is ${渭}_k=0.3$

The net force acting in vertical direction is zero as there is no motion.

$R-mg=0\left(1\right)$

As the static friction force is equal to and opposing the maximum conveyor belt force in the horizontal direction, the crate does not slip.

$$\begin{gathered} f_{max}=f_s \\ m{a}_{max}={渭}_{s}R \\ Fromequation\left(1\right) \\ m{a}_{max}={渭}_{s}mg \\ {a}_{max}={渭}_{s}g\left(II\right) \end{gathered}$$

Substitute the values where ${渭}_s=0.5,g=9.8m/s^2$

$$\begin{gathered} a_{max}=0.5脳9.8m/s^2 \\ {a}_{max}=4.9m/s^2 \end{gathered}$$

The maximum acceleration of the belt is $4.9m/s$