Answered

a 28 kg box rests on the back of a truck. the coefficient of static friction between the box and the truck bed is 0.334. the acceleration of gravity is 9.81 m/s2 . what maximum acceleration can the truck have before the box slides backward?



Answer :

The maximum acceleration will be 3.27 meter per second square.        

We have given that mass of the box m = 28 kg

Coefficient of static friction u = 0.334

Acceleration due to gravity g = 9.8m/s^2

For maximum acceleration

umg = ma

Cancelling m from both sides we get

ug = a

thus on substituting the values which we have mentioned above we get;

So acceleration will be 9.81 X 0.334 m/s^2

So maximum acceleration will be 3.27 m/s^2

Thus to know more about acceleration you may visit the link which is mentioned below:

https://brainly.com/question/12550364

#SPJ4

Other Questions