Calculating acceleration of a box from force?

AI Thread Summary
The discussion focuses on calculating the acceleration of a box on a truck based on the forces acting on it. The box has a mass of 15.0 kg, with static and kinetic friction coefficients of 0.8 and 0.63, respectively. The maximum acceleration of the truck before the box slides is determined to be 7.84 m/s², but the user struggles with calculating the box's acceleration once it begins to slide. They incorrectly apply the net force equation and misunderstand the forces acting on the box, leading to errors in their calculations. The maximum deceleration of the truck without the box sliding is also miscalculated, highlighting the need for clarification on the forces involved.
InertialRef
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Homework Statement



A box rests on top of a flat bed truck. The box has a mass of m = 15.0 kg. The coefficient of static friction between the box and truck is μs = 0.8 and the coefficient of kinetic friction between the box and truck is μk = 0.63.

I had previously calculated in an earlier part of the question that the maximum acceleration acceleration the truck can have before the box starts to move is 7.84 m/s^2. When the truck is at that acceleration, and the box begins to slide. What is the acceleration of the box?

With the box still on the truck, the truck attains its maximum velocity. As the truck comes to a stop at the next stop light, what is the magnitude of the maximum deceleration the truck can have without the box sliding?

Homework Equations



F = ma
Ff = μN

The Attempt at a Solution



If the box is no longer at rest on the back of the truck, then the net force is no longer equal to zero. From the free body diagram that I drew, there are two forces acting upon the box. The forward force of the box as a result of the acceleration of the truck and the force of friction in the opposite direction.

Hence: Fnet = Fforward + Ffriction

The forward force would just be the mass of the box times the acceleration of the truck, and the Ffricton = 0.63 * 15.0 * 9.81. Since friction is in the opposite direction, the direction of the force would be negative. So,

Fnet = (15.0)(7.84) + [-(0.63)(15.0)(9.81)]
ma = 117.6 - 92.70
(15)a = 24.89
a = 24.89/15
a = 1.6597 m/s^s

However, this answer is wrong.

As for the second part of that question, the maximum negative acceleration of the truck should be simply -7.84 m/s^2, since that is the maximum positive acceleration the truck can have without the box sliding. This is wrong as well.

If someone could tell me where in my reasoning I'm making the mistake, that would be very much appreciated. :)
 
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InertialRef said:
If the box is no longer at rest on the back of the truck, then the net force is no longer equal to zero. From the free body diagram that I drew, there are two forces acting upon the box. The forward force of the box as a result of the acceleration of the truck and the force of friction in the opposite direction.

Hence: Fnet = Fforward + Ffriction
There's only one (horizontal) force acting on the box--the friction from the truck bed.

As for the second part of that question, the maximum negative acceleration of the truck should be simply -7.84 m/s^2, since that is the maximum positive acceleration the truck can have without the box sliding. This is wrong as well.
For one thing, they want the magnitude, which is always positive.
 
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