NEED CONCEPTUAL HELP ONLY!(adsbygoogle = window.adsbygoogle || []).push({});

A package of eggs (mass 60.0 kg) sits on the flatbed of a pickup truck with an open tailgate. The coefficient of static friction between the package and the truck's flatbed is 0.350, and the coefficient of kinetic friction is 0.250.

a) The truck accelerates forward on level ground. I found the maximum acceleration the truck can have without the eggs sliding off the truck bed is given by (.350)(9.8 m/s/s) = 3.43 m/s/s by setting [itex]\mu[/itex]s*mg = ma. Easy.

(b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the package sliding along its bed. What is the acceleration of the eggs relative to the ground?

I found this by setting [itex]\mu[/itex]k*mg = ma, which gives, canceling out m, (.250)(9.8 m/s/s) = 2.45 m/s/s, which is RIGHT.

Now, what I am having trouble understanding is what is meant in part (b) by acceleration of eggs relative to the ground. How is this different from the acceleration of the eggs with respect to the truck? What would the acceleration of the eggs relative to the truck be? And why, then if acceleration of the eggs is given just by a = [itex]\mu[/itex]k*g, why it is not in the negative direction? I understand the truck is a non-inertial reference frame, so if the truck were accelerating, what would then be the acceleration of the eggs on the truck sliding on the truck relative to the ground?

Help please!

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# Conceptual Help! acceleration of eggs sliding on truck using kinetic friction

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