Did I do this physics problem right?

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The coefficient of friction between the box and the floor is calculated using the formula F = μN, where F is the applied force and N is the normal force. In this scenario, a 10 kg box is pushed with a constant force of 25 N, resulting in a coefficient of friction (μ) value of approximately 0.2551. The problem confirms that since the box moves at a constant velocity, the net force acting on it is zero, validating the calculations presented.

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Its not for homework, I just want to make sure I can do this.

What is the value of the coefficient of friction between the box and the floor?
A man pushes a box with a mass of 10kg at a constant velocity across the floor. he pushes with a constant force of 25N. What is the coefficient of friction value?


F = ukN
uk = F / N
uk = (25) / ((10)(9.8))
uk = 0.25
 
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You will need to know at what rate the box accelerates at in order to find the friction co-efficient.
 
LostConjugate said:
You will need to know at what rate the box accelerates at in order to find the friction co-efficient.

Not for this problem. If the velocity is constant, which the problem stated, then the net force is zero.

So mgcosθμ = 25

98μ = 25

μ = .2551...

so yes you're correct.
 

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