What Is the Coefficient of Kinetic Friction in This Physics Problem?

AI Thread Summary
The problem involves pulling a 250N crate with a 50N force at constant velocity, leading to the calculation of the coefficient of kinetic friction. The frictional force is equated to the applied force, resulting in the equation 50N = coefficient x 250N. Dividing both sides gives a coefficient of kinetic friction of 0.2. The solution is confirmed as correct, indicating that the calculations are straightforward. Understanding this concept is essential for solving similar physics problems.
Spartan Erik
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Homework Statement


"A forward horizontal force of 50N is used to pull a crate, which weighs 250N, at a constant velocity across a horizontal floor. The coefficient of kinetic friction is:"

Homework Equations


Frictional force = coefficient of kinetic friction x normal force

The Attempt at a Solution


This problem seems too easy, that's why I'm wondering if I might be wrong:

Frictional force = coefficient of kinetic friction x normal force
50N = coefficient x 250N
50N / 250N = coefficient
coefficient = 0.2

Is that all? Thanks
 
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yup that is correct.
 
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