Coefficient of kinetic friction?

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction for a 2.79-kg block on a 28.1° incline, launched at 3.87 m/s and sliding 0.969 m before stopping. The key equations used include Newton's second law in both x and y directions, specifically -ukN - mgsin(theta) = max and N = mgcos(theta). The final calculated coefficient of kinetic friction is 0.360, after correcting for calculator settings and ensuring proper sign conventions in the equations.

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juju1

Homework Statement


A 2.79-kg block is placed on a rough surface inclined at 28.1°. The block is launched at a speed of 3.87 m/s up the incline, and comes to rest after sliding 0.969 m up the incline.

Find the coefficient of kinetic friction between the block and the incline.

Homework Equations

The Attempt at a Solution


[/B]
II drew a diagram first, used a coordinate system with the x axis pointed up the slope and the y axis perpendicular to the slope. Got equations in both the X and Y direction.
x: -f-mgsin(theta)=ma_x
y: N-mgcos(theta)=0

But now I'm confused of what to do...
 
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Since the forces are constant, the acceleration is constant. Do you know any equations for constant acceleration that might help determine the value of ax?

Can you state a relevant equation that relates f and N?

(Your F = ma equations for x and y look good.)
 
v^2=v0^2+2a(delta x)?

i don't know how to write equations on here..ahaha.
 
f = coefficient of u_k N
and N is equal to mgcos(theta)
 
OK. So far, looks good. Go for it.
 
i got 1.42..which seems really large for coefficient of kinetic friction
 
juju1 said:
i got 1.42..which seems really large for coefficient of kinetic friction
Should ax be a positive number or a negative number when you substitute it into your x equation?
 
i got ax to be negative
 
juju1 said:
i got ax to be negative
Good. Can you type out the steps of your calculation of f using your x equation. I suspect you are making a sign error.
 
  • #10
x= -f-mgsin(theta)=max

-ukN-mgsin(theta)=max

-ukmgcos(theta)-mgsin(theta)=max

-ukmgcos(theta)=ma+mgsin(theta)

uk=(max+mgsin(theta))/(-mgcos(theta)
 
  • #11
juju1 said:
x= -f-mgsin(theta)=max

-ukN-mgsin(theta)=max

-ukmgcos(theta)-mgsin(theta)=max

-ukmgcos(theta)=ma+mgsin(theta)

uk=(max+mgsin(theta))/(-mgcos(theta)
That all looks good. Plug in the numbers, being careful with the signs, and see what you get.
 
  • #12
-0.462 ?
 
  • #13
juju1 said:

Homework Statement


A 2.79-kg block is placed on a rough surface inclined at 28.1°. The block is launched at a speed of 3.87 m/s up the incline, and comes to rest after sliding 0.969 m up the incline.

Find the coefficient of kinetic friction between the block and the incline.

Homework Equations

The Attempt at a Solution


[/B]
II drew a diagram first, used a coordinate system with the x axis pointed up the slope and the y axis perpendicular to the slope. Got equations in both the X and Y direction.
x: -f-mgsin(theta)=ma_x
y: N-mgcos(theta)=0

But now I'm confused of what to do...
You can also solve this problem by applying work energy theorem easily
 
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  • #14
juju1 said:
-0.462 ?
No. The coefficient should be positive (and I don't believe it is 0.462).

Can you show the last equation in your post #10 with all the numbers plugged in? Don't forget that g represents the magnitude of the acceleration due to gravity and so g is a positive number.
 
  • #15
( (2.79 x -7.73) + (2.79 x 9.8 x sin(28.1) ) / -(2.79 x 9.8 cos(28.1)
 
  • #16
juju1 said:
( (2.79 x -7.73) + (2.79 x 9.8 x sin(28.1) ) / -(2.79 x 9.8 cos(28.1)
That all looks good. Are you still getting the same answer as you got before?
 
  • #17
Yes.. :(
 
  • #18
I get an answer that is less than 0.4 when I use my calculator to evaluate your expression in post #16. Make sure your calculator is in "degree mode" rather than "radian mode".

What do you get for the overall numerator? For the denominator?
 
  • #19
My calculator was in radians...ahah! I got 0.360 :-)
 
  • #20
Thank you very much for your help!
 
  • #21
OK. That looks right. Good work!
 
  • #22
Abhishek kumar said:
You can also solve this problem by applying work energy theorem easily

I don't think the work energy principle is what is being taught here. It's just kinematics and forces, and in this situation the easiest.
 

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