To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.
Ff = 0.5 * 16 * 9.8 = 78.4 N
a = 4.9 m/s^2
If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?
In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.
Ff = 0.28 * 16 * 9.8 = 43.904 N
Net force = 78.4 – 43.904 = 34.496 N
To determine the acceleration, divide by the mass of the crate.
a = 34.496 ÷ 16 = 2.156 m/s^2
Explanation:
Formula to determine the critical crack is as follows.

= 1,
= 24.1
[/tex]\sigma_{y}[/tex] = 570
and, 
= 427.5
Hence, we will calculate the critical crack length as follows.
a = 
= 
= 
Therefore, largest size is as follows.
Largest size = 2a
= 
= 
Thus, we can conclude that the critical crack length for a through crack contained within the given plate is
.
The answer is A) the sand turns to liquid
Use the impulse-momentum theorem.

Substitute your known values:

Hope this helps!