Wood-block Friction Launch Problem

[Shaku]

Homework Statement

A 2.0kg wood block is launched up a wooden ramp that is inclined at a 25° angle. The block's initial speed is 9.0m/s . The coefficient of kinetic friction of wood on wood is μk = 0.200.

a) What vertical height does the block reach above its starting point?
b) What speed does it have when it slides back down to its starting point?

Homework Equations

(solve this without using the conservation of energy equation -- the 1/2*m*v^2 one because we haven't gone over that in class yet)

Vf² = Vi²+as

The Attempt at a Solution

I tried a lot of different things, but my numbers just wouldn't work out.

I put my Y axis along the ramp, and my X axis perpendicular to the ramp.

For part a:
I know that I have to:
1) Find the acceleration (mgCos$\theta$ for X direction and mgSin$\theta$ for Y direction)

2) Somehow incorporate the friction into the acceleration (I have no idea how to do this... We haven't gone over anything like this in class yet, that's why I'm so confused on this extra credit problem!)

3) Plug the acceleration I found into the Vf² equation to find s (height).

For part b:
I'd have to find the acceleration going downward incorporating the friction. But, again, I have absolutely no idea how to use the friction...My work so far:
I already have my axis set up and, for the most part, figured out. I'm still getting the wrong answer though.

Given:
2kg = 2kg(9.8) = 19.6N weight
Angle = 25°
Vi = 9m/s
μ_k = .2

X-axis is parallel to the ramp, Y axis is perpendicular to the ramp. Find the height (which is in the X direction).

Find a_x:
Fg_x + F_k = ma_x
mgSin∅ + mM_kCos∅ = ma_x
gSin∅ + M_kgCos∅ = a
g(Sin∅ + M_kCos∅) = a

9.8(Sin(25) + (9.8)(.2)(Cos(25)) = a
a = 5.918

Plug into Vf² = Vi² + ah
0 = 9² + (5.918)h
13.68 = h

But it said this answer for height was wrong... I'm not sure how to fix what I've done so far since to me it seems like it should work.

 

[tiny-tim]

Hi Shaku!

2) Somehow incorporate the friction into the acceleration (I have no idea how to do this...

add (or, in this case, subtract) the two forces along the slope …

mgsinθ and the friction force ​

(and then of course divide by mass to get the acceleration: F_total = ma)

 

[Simon Bridge]

Draw the free body diagram and put $\sum F = ma$ The lynchpin is to point your x-axis along the slope.

Friction is $f=\mu N$ pointing opposite the direction of motion. $N$ is the normal force to the surface the friction is against. If the surface was horizontal, then $f=\mu m g$
Encorporate it as another force on your free body diagram.

 

[Shaku]

Hi Shaku!


add (or, in this case, subtract) the two forces along the slope …

mgsinθ and the friction force ​

(and then of course divide by mass to get the acceleration: F_total = ma)

I need help on part a before I worry about part b. :P

 

[Shaku]

I already have my axis set up and, for the most part, figured out. I'm still getting the wrong answer though.

Given:
2kg = 2kg(9.8) = 19.6N weight
Angle = 25°
Vi = 9m/s
μ_k = .2

X-axis is parallel to the ramp, Y axis is perpendicular to the ramp. Find the height (which is in the X direction).

Find a_x:
Fg_x + F_k = ma_x
mgSin∅ + M_kCos∅ = ma_x
gSin∅ + M_kgCos∅ = a
g(Sin∅ + M_kCos∅) = a

9.8(Sin(25) + .2(Cos(25)) = a
a = 5.198

Plug into Vf² = Vi² + ah
0 = 9² + (5.918)h
13.68 = h

But it said this answer for height was wrong... I'm not sure how to fix what I've done so far since to me it seems like it should work.

 

[Simon Bridge]

I already have my axis set up and, for the most part, figured out. I'm still getting the wrong answer though.

Given:
2kg = 2kg(9.8) = 19.6N weight
Angle = 25°
Vi = 9m/s
μ_k = .2

X-axis is parallel to the ramp, Y axis is perpendicular to the ramp. Find the height (which is in the X direction).

Find a_x:
Fg_x + F_k = ma_x
mgSin∅ + M_kCos∅ = ma_x
gSin∅ + M_kgCos∅ = a
g(Sin∅ + M_kCos∅) = a

9.8(Sin(25) + .2(Cos(25)) = a
a = 5.198

Plug into Vf² = Vi² + ah
0 = 9² + (5.918)h
13.68 = h

But it said this answer for height was wrong... I'm not sure how to fix what I've done so far since to me it seems like it should work.

That's because you just calculated the distance the block went in the x direction - along the ramp. You need the height. Take another look at you diagram - how do you get the height from a direction in x?

 

[Shaku]

That's because you just calculated the distance the block went in the x direction - along the ramp. You need the height. Take another look at you diagram - how do you get the height from a direction in x?

Wouldn't it just be: 13.68*Sin(25)? I tried that too, but it still said it was wrong.

 

[Simon Bridge]

Then you need to look harder at your working:

mgSin∅ + MkCos∅ = ma_x
gSin∅ + MkgCos∅ = a

... how did you get from the top line to the next one?

9.8(Sin(25) + .2(Cos(25)) = a
a = 5.198

Check your arithmetic too.

 

[Shaku]

Then you need to look harder at your working:... how did you get from the top line to the next one?
Check your arithmetic too.

mgSin∅ + mMkCos∅ = max
divide by m, you get:
gSin∅ + M_kgCos∅ = a

-9.8(Sin(25) + (-9.8).2(Cos(25)) = a
a = -5.918

Then, finding the distance along the ramp:
Vf² = Vi² + ah
0 = 9² - 5.918s (--distance along the ramp)
s = 81/5.918
s = 13.68

Then, finding the height:
13.68(Sin(25) = h
h = 5.78, which is the same as my previous answer...

I'm really lost...

 

[Shaku]

I found an example from my notes that is pretty similar to this question, and it had the equation:

gSin∅ + MkgCos∅ = a

as well, so I'm not sure where I'm going wrong. I fixed my arithmetic (which wasn't actually an error, I just didn't write it up on the post correctly), and I did the exact same steps as the notes to find the length of the ramp. -- After that, I just multiplied the length of the ramp by the Sin(angle) and I, theoretically, SHOULD have gotten the right answer.

I have no idea where I'm screwing this up.

 

[Simon Bridge]

what direction is the acceleration?

 

[Shaku]

what direction is the acceleration?

It's negative, but that wouldn't change much in my final answer. Just an extra minus sign out front, and it's saying that that's wrong too.

Edit: Actually, it wouldn't change the answer at all as I had already done the calculations as if they were with a negative acceleration (I just wrote it down wrong).

See:
Vf² = Vi² + ah
0 = 9² - 5.918s (--distance along the ramp)
s = 81/5.918
s = 13.68

Which is the same thing.

 

[Simon Bridge]

You have to look harder: don't expect other people to spot this stuff - especially not if your reporting is full of typos.

Vf² = Vi² + ah

... is that the correct equation? It is hard to follow your work because you keep changing the variable name without warning. If you do this in work you hand in for marking, then it is costing you marks.