How Does Friction Affect Acceleration in a Pushed Box?

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The discussion focuses on calculating the acceleration of a 2.0 kg box being pushed with a force of 5.0 N across a flat surface, considering a friction coefficient of 0.2. The static friction force is calculated as 3.924 N, leading to a net force of 1.076 N. Using Newton's second law, the acceleration is determined to be 0.538 m/s². The calculations confirm the correct application of physics principles in determining the box's acceleration.

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PiRho31416
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I just want to check to see if I did the problem correctly.

A person pushes a 2.0 kg box across a flat, horizontal surface with a force of 5.0 N for 4.0 m. Determine the acceleration of a box if the friction coefficient is 0.2.

Determine the acceleration of the box if the friction coefficient is 0.2.

First, we need to find the static force.
F_s= \mu F_n = (0.2) (2 \cdot 9.81) = 3.924 \ N

Then, we need to find the net force.
F_{net} = 5 \ N - 3.924 \ N = 1.076 \ N

Then, finally, the acceleration of the box is:
F=ma \Rightarrow 1.076 = (2 \ kg) \ a \Rightarrow \boxed{a=0.538 m/s^2}

Is that right? Any help is appreciated.
 
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The 5N force was horizontal?
Than that was just about perfect.

It is best practice to do the algebra before putting the numbers in.
So: From Newton's second law: $$F+\mu mg = ma \implies a=\frac{F+\mu mg}{m}$$... F=5N, ##\mu##=0.2, and m=2kg;$$a=0.538\text{m/s}^2$$
 
Simon,

Thanks so much! :)

Adrian
 

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