What is the maximum acceleration the belt can have without the crate slipping? express your answer using two significant figures?

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

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation