Question

A horizontal force of 300.0 N is required to pull a 145.0 kg trunk across the floor
at a constant speed. Find the coefficient of kinetic friction.

Answer 1

So, the coefficient of kinetic friction approximately 0.21.

Introduction

Hi ! Here I will help you to solve the problem of the coefficient of kinetic friction. An object that experiences kinetic friction, indicates that the object move from its place. Remember that friction is also a force, but reduces the value of the force acting. Kinetic friction force is influenced by the normal force (N) and the coefficient of kinetic friction. The greater the value of the normal force (N) and/or the coefficient of friction, the greater the value of the kinetic friction force. [See the attached image to see the breakdown of the friction force].

Formula Used

For kinetic friction, there are the following equations that you will use in various problem conditions. However, the equation in the box is the formula you use for solve this problem.

$\sf{\sum F = m \cdot a}$

$\sf{F - f_k = m \cdot a}$

$\boxed{\sf{\bold{F - \mu_k \cdot m \cdot g = m \cdot a}}}$

With the following condition:

• $\sf{\sum F}$ = resultant of the force (N)
• F = acting force (N)
• m = mass of the object (kg)
• a = acceleration (m/s²)
• $\sf{f_k}$ = kinetic friction (N)
• g = coefficient of the earth gravity ≈ 9.8 m/s²
• $\sf{\mu_k}$ = coefficient of the kinetic friction

Solution

We know that:

• F = acting force = 300.0 N
• m = mass of the object = 145.0 kg
• a = acceleration = 0 m/s² >> Even if the object is moving, if we see the word "constant velocity", its means the acceleration is 0 m/s²
• g = coefficient of the earth gravity ≈ 9.8 m/s²

What was asked ?

• $\sf{\mu_k}$ = coefficient of the kinetic friction = ... ?

Step by step:

$\sf{F - \mu_k \cdot m \cdot g = m \cdot a}$

$\sf{300 - \mu_k \cdot 145 \cdot 9.8 = 145 \cdot 0}$

$\sf{-\mu_k \cdot 1,421 = -300}$

$\sf{\mu_k = \frac{-300}{-1,421}}$

$\boxed{\sf{\mu_k \approx 0.21}}$

Conclusion

So, the coefficient of kinetic friction approximately 0.21.

Learn More

After learning about kinetic friction. Are you still remember about static friction? If not, visit

https://brainly.com/question/28353910