Calculating acceleration of a box from force?

Click For Summary
SUMMARY

The discussion centers on calculating the acceleration of a box resting on a truck bed, given its mass of 15.0 kg and coefficients of static and kinetic friction (μs = 0.8, μk = 0.63). The maximum acceleration of the truck before the box begins to slide is determined to be 7.84 m/s². The user incorrectly calculates the box's acceleration when sliding and the maximum deceleration of the truck, leading to confusion about the net forces acting on the box. The correct approach emphasizes that only the frictional force opposes the box's motion once it begins to slide.

PREREQUISITES
  • Understanding Newton's Second Law (F = ma)
  • Knowledge of friction coefficients (static and kinetic)
  • Ability to analyze free body diagrams
  • Basic algebra for solving equations
NEXT STEPS
  • Review the concept of net force and its application in dynamics
  • Study the differences between static and kinetic friction
  • Learn how to correctly interpret and draw free body diagrams
  • Explore advanced problems involving friction and acceleration
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of friction and acceleration problems.

InertialRef
Messages
24
Reaction score
0

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. :)
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
6
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 4 ·
Replies
4
Views
17K
  • · Replies 11 ·
Replies
11
Views
4K