Solving Homework: Man Pushes 11.2 kg Box

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To solve the problem of a man pushing an 11.2 kg box at a constant speed of 3.5 m/s² on a horizontal surface with a coefficient of friction (μ) of 0.20, the horizontal force required is calculated using F=ma and F=μN. The applied force to maintain motion is determined to be 61.152 N, combining the force needed to overcome friction (21.95 N) and the force due to acceleration (39.2 N). For part b, if the pushing force is removed, the box will slide until it comes to rest, with the frictional force of 21.95 N acting against its motion. The discussion raises a question about whether the box is moving at constant speed or accelerating, which is crucial for accurately determining the forces involved. Understanding the distinction between constant speed and acceleration is essential for solving the problem correctly.
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Homework Statement


Man pushes a box that is 11.2 kg on a horizontal surface at a constant speed of 3.5 m/s2. Where μ is .20
a)What is the horizontal force rthat teh man must apply to maintain teh motion
b)If the force of part a is removed, how far does the box slide before coming to rest?

Homework Equations


F=ma
F=μN


The Attempt at a Solution


F=ma
F=11.2(3.5)
F=39.2

then
F=μN
F=(.20)(109.76)... [11.5*9.8]
F=21.95

then to get the total force i added 39.2+21.95 and got a total of 61.152

For part b I would asume that without part a the force would just be 21.95 because it doesn't account for the answerr to part a?
 
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Man pushes a box that is 11.2 kg on a horizontal surface at a constant speed of 3.5 m/s2

Constant speed or acceleration?
 
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