Coefficient of kinetic friction?

[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...

 

[TSny]

Since the forces are constant, the acceleration is constant. Do you know any equations for constant acceleration that might help determine the value of a_x?

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

(Your F = ma equations for x and y look good.)

 

[juju1]

v^2=v0^2+2a(delta x)?

i don't know how to write equations on here..ahaha.

 

[juju1]

f = coefficient of u_k N
and N is equal to mgcos(theta)

 

[TSny]

OK. So far, looks good. Go for it.

 

[juju1]

i got 1.42..which seems really large for coefficient of kinetic friction

 

[TSny]

i got 1.42..which seems really large for coefficient of kinetic friction

Should a_x be a positive number or a negative number when you substitute it into your x equation?

 

[juju1]

i got a_x to be negative

 

[TSny]

i got a_x 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.

 

[juju1]

x= -f-mgsin(theta)=ma_x

-u_kN-mgsin(theta)=ma_x

-u_kmgcos(theta)-mgsin(theta)=ma_x

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

u_k=(ma_x+mgsin(theta))/(-mgcos(theta)

 

[TSny]

x= -f-mgsin(theta)=ma_x

-u_kN-mgsin(theta)=ma_x

-u_kmgcos(theta)-mgsin(theta)=ma_x

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

u_k=(ma_x+mgsin(theta))/(-mgcos(theta)

That all looks good. Plug in the numbers, being careful with the signs, and see what you get.

 

[juju1]

-0.462 ?

 

[Abhishek kumar]

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

 

[TSny]

-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.

 

[juju1]

( (2.79 x -7.73) + (2.79 x 9.8 x sin(28.1) ) / -(2.79 x 9.8 cos(28.1)

 

[TSny]

( (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?

 

[juju1]

Yes.. :(

 

[TSny]

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?

 

[juju1]

My calculator was in radians...ahah! I got 0.360 :-)

 

[juju1]

Thank you very much for your help!

 

[TSny]

OK. That looks right. Good work!

 

[lekh2003]

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.