A wood block is launched up a ramp

[kenway]

Homework Statement

A 3 kg wood block is launched up an inclined wooden ramp that has an angle of 25° to the horizontal. the Block's initial speed is 12 m/s and the kinetic coefficient of friction is 0.2. What is the vertical height the block reaches above it's starting point? What speed does it have when it slides back down to its starting point?

Homework Equations

The Attempt at a Solution

What's listed first is what is given:

V_O=12m/s
g=9.8m/s²
mass (m) = 3kg, which means
weight (w) = 29.4 N
μ=0.2
θ=25°

Then I found the x and y components of the weight:
w_x = 29.4sin25 = 12.42 N
w_y = 29.4cos25 = 26.65 N

The normal force and y component of the weight equal each other:
normal force (n) = w_y
n=26.65 N

And by using the normal weight I found friction:
friction (f) = 0.2(26.65N) (μ*n)
f= 5.33 N

This is where I got stuck. I know if I find the acceleration in the x-direction of the block I can calculate the distance it traveled, which would give me the hyp. of the ramp (triangle) and I can use trig to find the y-comp. So if I find acceleration:
a = (w_x-f)/m
a = (12.42 - 5.33)/3
a=2.63

This doesn't help because it should be negative. But then I read somewhere gravity is also acting as a resisting force, so then I modified my equation:

a = (12.42-5.33-9.8)/3
a = -0.90

But this seems low. I've tried altering that equation other times but I feel every time it is wrong. Could I get some guidance?

 

[TSny]

Hello. Welcome to PF.

This is where I got stuck... So if I find acceleration:
a = (w_x-f)/m

Consider the directions of w_x and f.

 

[kenway]

They're both negative? So I would set it up as:

a = (-12.42 + (-5.33))/3
a = -5.92 m/s²

I wouldn't subtract the friction, because the two forces are adding on each other, so to say.

Then I can find time:

v = at
12/-5.92=t
2.03 s = t

Then I can plug all my variables into this kinematic equation:

x = x₀ + v_x0t + .5a_xt²
x = 0 + 0(2.03) + .5(-5.92)(2.03²)
x = -12.20 m

This should give me the distance traveled by the block, or the hyp. of the triangle, but it's negative and that doesn't make sense. I'm not sure if the initial velocity (v_x0) should be 12 though. If I plug 12 in for that, then I just add 24.36 to -12.20 and get 12.16 m. If that is the answer, then I can use this equation to find the height:

12.16sin25=y
5.14 m = y

And supposedly this is my answer. Does this look right?

 

[TSny]

They're both negative? So I would set it up as:
a = (-12.42 + (-5.33))/3
a = -5.92 m/s²
I wouldn't subtract the friction, because the two forces are adding on each other, so to say.

Yes.

Then I can find time:
v = at
12/-5.92=t
2.03 s = t

OK

x = x₀ + v_x0t + .5a_xt²
x = 0 + 0(2.03) + .5(-5.92)(2.03²)
x = -12.20 m

This should give me the distance traveled by the block, or the hyp. of the triangle, but it's negative and that doesn't make sense.

Right, negative x indicates an error.

 

[kenway]

As I described in the above paragraph, I tried again by plugging in 12 for the 0 (that you have highlighted) and got 12.16 m. If that is the answer, then I can use this equation to find the height:

12.16sin25=y
5.14 m = y

I think this is the answer

 

[TSny]

As I described in the above paragraph, I tried again by plugging in 12 for the 0 (that you have highlighted) and got 12.16 m. If that is the answer, then I can use this equation to find the height:

12.16sin25=y
5.14 m = y

I think this is the answer

I believe that's correct.

 

[kenway]

So how do I calculate the second answer of the question: What speed does it have when it slides back down to its starting point?
I honestly don't even know where to start.

 

[cnh1995]

So how do I calculate the second answer of the question: What speed does it have when it slides back down to its starting point?
I honestly don't even know where to start.

The block starts sliding from rest. Now, which force will make it slide and which force will oppose the sliding?

 

[kenway]

Gravity causes it to slide, friction opposes it. So do I do it like this?:

a = (9.8-5.33)/3

Gravity - friction (found above) / mass

a = 1.49 m/s²

Then I would find the time it takes to get to the bottom? So:

y = .5at²
5.14 = .5(1.49)t²
sqrt(5.14/.0745 = t
2.63 s = t

Which I can then use in this equation:

v=at
v = 1.49(2.63)
v = 3.92 m/s

Does this look right?

 

[cnh1995]

Friction is 5.33N, fine. But gravity is acting "along the incline" .Would it be 9.8? That's only acceleration and that too in the downward direction and not along the incline.

 

[kenway]

Oh! So then I find the comp. that runs along the incline. So:

sin25=9.8/hyp
9.8/sin25=hyp
23.19

The hyp runs along the incline, but this seems like a high number. If it's supposed to be, then I will go with it, but when I plugged it into the equations I did in the last post I got 20.59 m/s for the final speed (I can do the math if you want to see), which seems high.

 

[cnh1995]

Oh! So then I find the comp. that runs along the incline. So:

sin25=9.8/hyp
9.8/sin25=hyp
23.19

The hyp runs along the incline, but this seems like a high number. If it's supposed to be, then I will go with it, but when I plugged it into the equations I did in the last post I got 20.59 m/s for the final speed (I can do the math if you want to see), which seems high.

In the first part, component of gravitational force was resisting the block's motion by pulling it backward. That "same" force will now be causing it to slide. You've already calculated that in the first part, haven't you? And what you are taking as 9.8 is only the acceleration, not force as you haven't involved mass here.

 

[kenway]

So that would be the x-comp. of the weight? I found:

mgsinθ = x

But that was the x-comp. of the block's weight, not gravity. If that's not the correct answer, then I did the first part wrong

 

[TSny]

Going up, the block had two forces acting along the incline and they acted in the same direction.

Going down, what two forces act on the block along the incline? What are their directions? Get the acceleration using the same method as you did in the first part.

 

[kenway]

Going up, the block faced friction and it's own weight, but it's initial velocity overcame that.
Going down, it doesn't have the velocity, but the friction is now facing up, away from the direction the block is moving, and it's "weight" still faces down, which pushes it down the ramp, so to say. So:

a=(12.42-5.33)/3
a=2.36 m/s²

Plugging it into the equations, I got 4.91 m/s.

 

[cnh1995]

So that would be the x-comp. of the weight? I found:

mgsinθ = x

But that was the x-comp. of the block's weight, not gravity. If that's not the correct answer, then I did the first part wrong

Weight is a due to the gravity, isn't it?
The component of wight along the incline was resisting the block's motion in the first part. You did it right. Now the same force will make the block slide.

 

[cnh1995]

Going up, the block faced friction and it's own weight, but it's initial velocity overcame that.
Going down, it doesn't have the velocity, but the friction is now facing up, away from the direction the block is moving, and it's "weight" still faces down, which pushes it down the ramp, so to say. So:

a=(12.42-5.33)/3
a=2.36 m/s²

Plugging it into the equations, I got 4.91 m/s.

Your analysis is correct!

 

[kenway]

Okay, this makes more sense now! Thank you both for all your help. I feel much more confident about this. I appreciate it very much!

 

[TSny]

Going up, the block faced friction and it's own weight, but it's initial velocity overcame that.
Going down, it doesn't have the velocity, but the friction is now facing up, away from the direction the block is moving, and it's "weight" still faces down, which pushes it down the ramp, so to say. So:

a=(12.42-5.33)/3
a=2.36 m/s²

OK. ( I guess you are now taking the positive x direction to be down the incline.)

Plugging it into the equations, I got 4.91 m/s.

Hmm. Did you use the correct distance of travel?

 

[kenway]

I used the y comp. of the ramp (that's what I used in the first equation, 5.14m), but I can see where that could be wrong. The length of the ramp (hyp of the triangle) would then be 12.16 m. Then to find the time:

sqrt(12.16/2.36/2) = t
3.21 s = t

v=at
v = 2.36(3.21)
v = 7.58 m/s

 

[TSny]

Looks good.

 

[cnh1995]

I used the y comp. of the ramp (that's what I used in the first equation, 5.14m), but I can see where that could be wrong. The length of the ramp (hyp of the triangle) would then be 12.16 m. Then to find the time:

sqrt(12.16/2.36/2) = t
3.21 s = t

v=at
v = 2.36(3.21)
v = 7.58 m/s

Right! That can also be done without involving time t. You can use,
v²=u²+2ax..
You have u=0, a=2.36, x=12.16m.