Find Kinetic Friction of a Block Pushed Up an Incline

[muddy510]

1.
A block is projected up an incline making an angle theta with the horizontal. It returns to
its initial position with half its initial speed. Show that the coefficient of kinetic friction is
μk = (3/5)tan(theta)

(For a 'properly formatted' version of the question, see number 3 on the attached .pdf)2.
Not entirely sure. I think it may be possible to solve this using work-energy equations or kinetics.3.
Normal Force is mgcos(theta)
How should I properly start this problem?
Thank you for your consideration.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

welcome to pf!

hi muddy51! welcome to pf!

(have a theta: θ and try using the X² and X₂ icons just above the Reply box )

yes, use the work energy theorem …

call the height h, and the initial speed v, and do two work-energy equations, one for the way up, one for the way down …

what do you get?

 

[PhanthomJay]

One way to start it is to look first at the motion up the plane, then look next at the motion down the plane. For each case, use the equation noting the fact that the work done by friction is equal to the change in the mechanical energy of the system. Then solve the 2 equations to determine the friction coefficient. You'll get a bunch of terms that you can cancel out.

 

[muddy510]

Thank you both for the guidance. I was able to solve the problem. Is it policy that I post the solution?

 

[tiny-tim]

no! (unless you're still worried about something)

see you around! ​

 

[Rayquesto]

the kinetic energy back where it started is half the potential energy at its projected hieght. so, if mgcostheta is the normal force then the force down below would be half that right? if cos of 30 is Square root of 3 divided by 2 then half that is the square root of three which is the tangent of 60. any other ideas?

 

[Rayquesto]

I meant the force applied is half of the potential energy and the gravitational force at the maximum height.

 

[Rayquesto]

if we draw it out then the vector representing the applied 1/2 force would be 1/2 of the downward vector, therefor, finding applied force would be 1/2 of the gravitational force at maximum height which would be cos theta times mass times 9.81. 1/2 that would be the applied force, which is somehow involved in tangent. You have any ideas for this?

 

[penguin87]

Hi I'm new to threads I came across a similar question in my homework problems and am still unsure on how to solve it! I unerstand you need to set up two equations for up and down the hill but how do you incorperate the angle into the equations? Any help would be appreciated I'm a little lost!

 

[tiny-tim]

welcome to pf!

hi penguin87! welcome to pf!

I unerstand you need to set up two equations for up and down the hill but how do you incorperate the angle into the equations?

you need θ to find the normal https://www.physicsforums.com/library.php?do=view_item&itemid=73"

(and then you can multiply that by µ to find the https://www.physicsforums.com/library.php?do=view_item&itemid=39" force) …

what do you get? ​

 

[penguin87]

oh okay i understand that i just don't quite understand what my equations would look like. I'm pretty sure i have to use force calculations but not sure how to put it all together

 

[tiny-tim]

show us how far you've got, and where you're stuck, and then we'll know how to help!

 

[penguin87]

alright this is what i got so far!
going up the hill: Fnetx=Fapplied-Ffriction-Fgsinθ
Fnety=Fn-Fgcos
going down the hill: Fnetx=Fapplied-Ffriction+Fgsin
Fnety=Fn-Fgcos

i tried islating for the force of friction but i ended up getting a big mess of equations!
am i going in the right direction, and guidance would help:)

 

[tiny-tim]

hi penguin87!

(just got up :zzz: …)

going up the hill: Fnetx=Fapplied-Ffriction-Fgsinθ
Fnety=Fn-Fgcos
going down the hill: Fnetx=Fapplied-Ffriction+Fgsin
Fnety=Fn-Fgcos

(try using the X₂ tag just above the Reply box )

yes, that's ok so far, except that F_applied is zero …

after the initial https://www.physicsforums.com/library.php?do=view_item&itemid=340" (shove), there's no applied force

 

[penguin87]

Thank you so much that was throwing off my equation completely!
so i now solved that F_n=F_gcosθ
and when i solved for F_k i got -sinθ

my final answer for μ= -tanθ
does that make sense?

 

[tiny-tim]

hi penguin87!

so i now solved that F_n=F_gcosθ
and when i solved for F_k i got -sinθ

what's F_k ?

my final answer for μ= -tanθ
does that make sense?

nooo … i think you've put F_net = 0

F_net will be negative (both uphill and down hill), won't it?

try again ​

 

[penguin87]

sorry i made the force of kinetic friction equal to Fk!
and i made Fnet = -ma in both directions

then i turned a=v/t and velocity in the downwards direction was 1/2 of that in the upward direction!

it doesn't seem like my reasoning makes any sense and so I am getting the wrong answer

 

[tiny-tim]

It returns to
its initial position with half its initial speed.

sorry i made the force of kinetic friction equal to Fk!
and i made Fnet = -ma in both directions

then i turned a=v/t and velocity in the downwards direction was 1/2 of that in the upward direction!

it doesn't seem like my reasoning makes any sense and so I am getting the wrong answer

sorry, i don't follow what you've done

you should be getting a_up and a_down as a function of θ,

then you can use one of the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations to find the distance moved (ie when v = 0), then another equation to find v_final after the same distance down

alternatively, it would be easier to start as i said in post #2 …​

use the work energy theorem …

call the height h, and the initial speed v, and do two work-energy equations, one for the way up, one for the way down …

what do you get?