What is the coefficient of kinetic friction for a box pulled by a 10N force?

  • Thread starter Thread starter future_vet
  • Start date Start date
  • Tags Tags
    Box Force
Click For Summary
SUMMARY

The coefficient of kinetic friction for a 2.0 kg box being pulled by a 10N force with an acceleration of 0.50 m/s² is calculated using the equation F - μk mg = ma. By substituting the values into the equation, the correct coefficient of kinetic friction (μk) is determined to be 0.46. The calculations confirm that the initial approach was correct, but the final arithmetic needed adjustment to arrive at the accurate result.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with the concept of friction and its coefficients
  • Basic algebra for solving equations
  • Knowledge of gravitational force calculations (mg)
NEXT STEPS
  • Study the derivation of the friction coefficient formula
  • Learn about different types of friction (static vs. kinetic)
  • Explore applications of Newton's laws in real-world scenarios
  • Practice solving problems involving forces and acceleration
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of applying Newton's laws and friction in problem-solving contexts.

future_vet
Messages
169
Reaction score
0

Homework Statement


A 2.0 kg box is pulled along a horizontal plank by a 10N force parallel to the direction of motion. The rate of acceleration is 0.50 m/s^2. Find the coefficient of kinetic friction.

I think we use the formula: coefficient of friction = F (so 10) / N.
But I am not sure how to use the acceleration and the N variable.


Homework Equations


F - μk mg = ma
10 - 2(9.8)μk = 2 x 0.5

Would this be the correct way to go?

Thanks!
 
Physics news on Phys.org
future_vet said:

Homework Statement


A 2.0 kg box is pulled along a horizontal plank by a 10N force parallel to the direction of motion. The rate of acceleration is 0.50 m/s^2. Find the coefficient of kinetic friction.

I think we use the formula: coefficient of friction = F (so 10) / N.
But I am not sure how to use the acceleration and the N variable.


Homework Equations


F - μk mg = ma
10 - 2(9.8)μk = 2 x 0.5

Would this be the correct way to go?

Thanks!
Yes, but it contradicts what you said in the problem statement.
 
Yes, so I use the second equation? Is the data correctly plugged in?
 
future_vet said:
Yes, so I use the second equation? Is the data correctly plugged in?

Yes it is.
 
I am getting -0.005 as the answer... does it sound right? Where should the acceleration be used?

Thanks for all the help! :)
 
future_vet said:
I am getting -0.005 as the answer... does it sound right? Where should the acceleration be used?

Thanks for all the help! :)
you have the correct equation, with the given acceleration entered correctly, but your math is off.
10 -19.6\mu_k = 2(.5) = 1
-19.6\mu_k = -9
\mu_k = ?
 
oops, it's 0.46, right?
 
future_vet said:
oops, it's 0.46, right?
yes, correct.
 

Similar threads

Static Friction Between a Box and the Floor
  • · Replies 11 ·
Replies
11
Views
3K
Truck carrying a box, friction problem
  • · Replies 10 ·
Replies
10
Views
5K
Can a 10N Force Overcome Friction to Move a 10kg Block?
  • · Replies 2 ·
Replies
2
Views
2K
Calculating the coefficient of kinetic friction
  • · Replies 3 ·
Replies
3
Views
4K
Calculating acceleration from the coefficient of kinetic friction
  • · Replies 48 ·
2
Replies
48
Views
8K
Average Coefficient of Kinetic Friction between Ice and Puck
  • · Replies 3 ·
Replies
3
Views
4K
Forces Grade 12 Physics Question
  • · Replies 5 ·
Replies
5
Views
2K
Finding the force of kinetic friction without a coefficient
  • · Replies 8 ·
Replies
8
Views
18K
Why Does the Friction Force Not Match in the Two Block Problem?
  • · Replies 15 ·
Replies
15
Views
3K
Calculating the Minimum Force for Box to Move at Constant Velocity
  • · Replies 2 ·
Replies
2
Views
3K