How to find the coefficient of kinetic friction?

Click For Summary
SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction (μk) for a child sliding down a 28° incline, where her speed at the bottom is half of what it would be on a frictionless slide. The relevant equations include ΣF=ma, Ffr=μkFN, and the kinematic equation v²=v0²+2a(x−x0). The solution involves analyzing forces and energy, concluding that the kinetic energy with friction is one-fourth that of the frictionless case, leading to a definitive relationship between acceleration and friction.

PREREQUISITES
  • Understanding of Newton's second law (ΣF=ma)
  • Familiarity with kinetic energy concepts (KE=1/2mv²)
  • Knowledge of forces acting on an inclined plane (normal force and friction)
  • Basic algebra for solving equations involving variables
NEXT STEPS
  • Learn about energy conservation in mechanical systems
  • Study the relationship between acceleration and friction on inclined planes
  • Explore the derivation of kinetic energy equations in different scenarios
  • Investigate advanced friction models and their applications in real-world physics problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and friction, as well as educators seeking to explain the concepts of kinetic friction and energy conservation in inclined motion.

audrey1203
Messages
1
Reaction score
0

Homework Statement


A child slides down a slide with a 28° incline, and at the bottom her speed is precisely half what it would have been if the slide had been friction-less. Calculate the coefficient of kinetic friction between the slide and the child.
Knowns:
28°
speed=V/2

Homework Equations


##\Sigma F=ma##
##{ F }_{ fr }={ \mu }_{ k }{ F }_{ N }##
##{ v }^{ 2 }={ v }_{ 0 }^{ 2 }+2a(x-{ x }_{ 0 })##

The Attempt at a Solution


##\Sigma { F }_{ y }=ma=0##
##{ F }_{ N }-mgcos\sigma =0##
##{ F }_{ N }=mgcos\sigma##

##{ \Sigma F }_{ x }=ma##
##-{ F }_{ fr }+mgsin\sigma =ma##
##{ F }_{ fr }=\frac { ma-mgsin\sigma }{ -1 } ##
##{ F }_{ fr }=-ma+mgsin\sigma ##

##{ F }_{ fr }={ \mu }_{ k }{ F }_{ N }##
##-ma+mgsin\sigma ={ \mu }_{ k }mgcos\sigma ##
##\frac {-ma+mgsin\sigma }{ mgcos\sigma } ={ \mu }_{ k }##
##\frac { -a+gsin\sigma }{ gcos\sigma } ={ \mu }_{ k }##
 
Physics news on Phys.org
WELCOME TO PF!
audrey1203 said:
##-ma+mgsin\sigma ={ \mu }_{ k }mgcos\sigma ##
##\frac {-ma+mgsin\sigma }{ mgcos\sigma } ={ \mu }_{ k }##
##\frac { -a+gsin\sigma }{ gcos\sigma } ={ \mu }_{ k }##
OK. But you can see that solving for ##\mu_k## here does not seem to help much since you don't know the value of the acceleration ##a##.
Instead of using the first equation above to solve for ##\mu_k##, think of the equation as giving you the acceleration when friction is present.
What would this equation look like if there was no friction?

How must the acceleration with friction compare to the acceleration without friction in order for the final speed to be cut in half with friction?
 
Hi audrey1203, Welcome to Physics Forums!

One way to tackle the problem is to look at it in terms of kinematics: forces and accelerations, which is what you've shown. Another way is to consider an energy approach.

Since KE is proportional to the square of the velocity, and if the velocity with friction present is one half that of when there's no friction, then the KE with friction must be 1/4 that of the frictionless case. That is, ##KE_o = 4 KE_f##, where ##KE_o## is the kinetic energy of the child at the bottom of the slide when no friction is present , and ##KE_f## is the kinetic energy when friction is present. You should confirm that for yourself by writing out the expressions for KE when the velocity is v and v/2 and comparing the results.

Hint: If you choose an arbitrary value for the length of the slide (maybe call it L) then you can determine the change gravitational PE during the trip and any energy loss due to friction since you know how to find the friction force. All the unknowns (like L and the mass of the child) should disappear in the algebra along the route :smile:
 

Similar threads

Replies
11
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 21 ·
Replies
21
Views
1K
Replies
48
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 28 ·
Replies
28
Views
6K
Replies
7
Views
1K