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A person pushes a box of mass 15 kg along a smooth floor by applying a perfectly horizontal force F. In the process, the 15 kg box pushes against another box with a mass of 10 kg and causes it to move. The boxes accelerate horizontally at a rate of 2.0 m/s2.
Part A
What is the magnitude of F?
Solution
System: Both boxes together as a single point particle.
Interactions: External influences from the person (applied force) the earth (gravity) and the floor (normal force).
Model: Point Particle Dynamics.
Approach:
The word smooth in the problem statement is a keyword, telling us that the floor exerts no horizontal force on the box. Thus, the only external horizontal force acting on the system composed of the two boxes together is that applied by the person. This implies thatNewton's 2nd Law for the x direction can be written:
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\begin
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[ \sum F_
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= F = m_
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a_
= (\mbox
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+ \mbox
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)(\mbox
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^
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) = \mbox
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] \end
Part B
What is the force applied on the front (10 kg) box by the rear (15 kg) box?
Solution
System: This time, we will focus only on the front box.
Interactions: The front box is subject to external influences from the rear box, the earth (gravity) and the floor (normal force).
Model: Point Particle Dynamics
Approach:
For simplicity, we will refer to the front box as box 2 and the rear box as box 1. We will call the force applied by box 1 to box 2 F21. The only horizontal force experienced by the front box is the force F21 provided by the rear box. Thus, Newton's 2nd Law for the x direction can be written:
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\begin
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[ \sum F_
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= F_
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= m_
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a_
= (\mbox
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)(\mbox
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^
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) = \mbox
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] \end
Part C
Suppose we now consider the rear (15 kg) box as a
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