How to find the max. compression of a spring given a block on an incline?

[spacestrudel]

Homework Statement

A block is sliding down an inclined plane with friction. The block collides with a spring at the bottom of the incline. What is the max. compression of the spring?

Relevant Equations

Spring Potential Energy = 1/2kx^2. Ei = Ef.

Hello! I am stuck on part of a problem and was wondering what I am doing wrong. For part a of the problem, we were asked to find the impact speed. I did this in a photo below given the following values:

Θ = 30 degrees. The initial velocity = 10 m/s. The coefficient of kinetic friction = 0.4. distance between the spring and block = 5 meters. Mass of the block = 300kg. k = 20N/m.

My professor told us that when the spring is at max. compression, the velocity equals zero. So, I'm assuming we use that somehow?
Here is a photo of my work thus far (sorry it's blurry on here):

Initially I tried to perform the following calculations:
mg(hf + x) = 1/2kx^2
= 1/2kx^2 = mg(x+5)sin(30)
= (300)(9.81)(x+5)(sin(30)) = 100x^2
= (2,943 N)(x+5)(sin(30)) = 100x^2

Which eventually simplifies to the following:
(100)x^2 - (1471.5)x - 7357.5 = 0.

I plugged these values into the quadratic formula and received x = 18.658 meters. However, this seems far too large to be the maximum compression considering there is friction and I'm not sure how to take that into account. Should I perhaps multiply the friction times sin(30)? Or maybe just scrap this method and try a different set of equations?

Thanks so much!

 

[mfb]

Friction will take out energy of your system. As that energy is proportional to the mass it doesn't make the calculation more complicated.
What is the force of friction along the direction of motion?

 

[spacestrudel]

Friction will take out energy of your system. As that energy is proportional to the mass it doesn't make the calculation more complicated.
What is the force of friction along the direction of motion?

The force of friction I found as FE = u_kFNd (where uk represents the coefficient of kinetic friction). By plugging in the normal force and x for d, I then found this to be 2u_kgcos(Θ)x. (At least from above in terms of the final velocity calculations I performed for part a). And I am given that u_k = 0.4

Not sure how to apply this part of the problem to part b (or if it even carries over).

 

[mfb]

I'm not sure what part (a) or part (b) are.
How did you get 2u_kgcos(Θ)x? The distance is not proportional to x and I don't understand where the 2 comes from. There is also the mass of the block missing.

 

[spacestrudel]

Friction will take out energy of your system. As that energy is proportional to the mass it doesn't make the calculation more complicated.
What is the force of friction along the direction of motion?

So sorry let me clarify - part a) is the scanned image included in this post. The question asks, "what is the impact velocity?" So, by solving using conservation of energy, as well as replacing values such as the fact that FE can be replaced to equal u_kFNd which simply replaces to FN = mgcos(Θ) and d replaces to x. The 2 comes from the fact that to solve for final velocity I multiplied the entire equation by 2.
So, basically, I'm not sure if I used any of the information from part a) to solve part b) which is asking the max. compression of the spring.

 

[haruspex]

Here is a photo of my work thus far (sorry it's blurry on here):

The image is clear but your handwriting sometimes is not. Also, it is hard to refer to specific steps in an image. This is why forum rules say that images are only for diagrams and textbook extracts.

Something wrong with your signs in part a. Gravity seems to be slowing the block down instead of speeding it up.

For part b, I see nothing in the problem statement about friction ceasing as soon as the block touches the spring.

 

[spacestrudel]

The image is clear but your handwriting sometimes is not. Also, it is hard to refer to specific steps in an image. This is why forum rules say that images are only for diagrams and textbook extracts.

Something wrong with your signs in part a. Gravity seems to be slowing the block down instead of speeding it up.

For part b, I see nothing in the problem statement about friction ceasing as soon as the block touches the spring.

Ah, sorry about that I will write out my steps for part a. Also, after redoing it again at a reasonable hour instead of at 2 am, I got a different answer for Vf.

So, since part a) asks for the final velocity of the block right before impact, I did the following steps:
KE_i + ug₀ = KE_f + ug_f + FE. Where KE_i = initial kinetic energy, Ug₀ = initial gravitational potential energy, KE_f = final kinetic energy, Ug_f = final gravitational potential energy, and FE equals the potential energy of friction.

Since I said the height initial (at the top of the incline) equals zero, the initial gravitational potential energy cancels out since it is zero.

Furthermore, I'll say that d for distance simply equals x, FN = mgcos(30), FE = ukFNd --> which becomes uk(mgcos(30))x. Since I said that d = x and FN = mgcos(30).

Additionally, h_f (since it is at the bottom of the incline) equals xsin(30).

My equation now looks like the following (where v_i is the initial velocity and v_f is the final velocity. I got these from the fact that KE = 1/2mv^2):
1/2m*v^2_i = 1/2m*v^2_f + m*g*(xsin(30)) + u_k*m*x*gcos(30)

I cancel out all of the m's since each side has one and multiply the entire equation by 2.

It now is:

v^2_i - 2*g*x*sin(30) - 2u_k*x*g*cos(30) = v^2_f

Taking the square root of this all, vf now becomes: sqrt((v^2_i -2*g*x((sin(30)+ukcos(30)))

I find the answer of V_f to be 4.119 m/s.
_________________________________________________________________

Then, for part b) of the problem I calculate the max. compression of the spring by the following equation:

1/2m*v^2f + mghf = FE + 1/2kx^2

However, since I said h_f = xsin(30), what would the new final height be since we have to take into account the spring gets compressed even more than this?

 

[mfb]

You use x both for the distance until the mass hits the spring and the distance the spring compresses, this is confusing.

 

[haruspex]

Since I said the height initial (at the top of the incline) equals zero

OK, but what then is the sign of the final height? You still have a final equation that says gravity is acting to slow it down (v_i^2 - 2*g*x*sin(30) ...). Try to develop the habit of sanity-checking equations. Sign errors are annoyingly easy to make.

1/2m*v^2f + mghf = FE + 1/2kx^2

As @mfb posted, it is most confusing to reuse labels x, h_f with different meanings. If the spring is compressed by distance s, what is the vertical displacement?

 

[spacestrudel]

OK, but what then is the sign of the final height? You still have a final equation that says gravity is acting to slow it down (v_i^2 - 2*g*x*sin(30) ...). Try to develop the habit of sanity-checking equations. Sign errors are annoyingly easy to make.

As @mfb posted, it is most confusing to reuse labels x, h_f with different meanings. If the spring is compressed by distance s, what is the vertical displacement?

Ah, I see. So, I'm assuming it should be v_i^2 +2*g*x*sin(30) ...) since gravity would be speeding the block up. Which would then change my value of v_f.

So, if the spring is compressed by s, could I write that h_f = (s+x)sin(30) or something along those lines? (We know that x = 5 meters from the given above).

 

[haruspex]

it should be vi^2 +2*g*x*sin(30) ...) since gravity would be speeding the block up

Yes, but you should go back and understand where your sign error came from.

if the spring is compressed by s, could I write that hf = (s+x)sin(30)

Yes, but remember here you are only concerned with the additional descent after contacting the spring.
What about the FE term for that additional descent?

 

[spacestrudel]

Yes, but you should go back and understand where your sign error came from.

Yes, but remember here you are only concerned with the additional descent after contacting the spring.
What about the FE term for that additional descent?

I believe it stems from the fact that I should actually say hf = -xsin(30), since it's moving down on an incline?

And do I have to add s to the FE side, as well?

 

[haruspex]

it stems from the fact that I should actually say hf = -xsin(30), since it's moving down on an incline?

Yes.

do I have to add s to the FE side, as well?

Yes.

 

[spacestrudel]

Yes.

Yes.

Thank you! I received a rather small number for the max. compression (1.914 meters) of the spring, but I'm assuming this makes sense since the spring constant was only 20 N/M and the block had only traveled a bit more than 5 meters at this point. The final velocity of the block that I calculated correctly (finally) was 10.727 meters. The block's mass was 300kg which led me to believe that the spring's compression might be more?

 

[haruspex]

Thank you! I received a rather small number for the max. compression (1.914 meters) of the spring, but I'm assuming this makes sense since the spring constant was only 20 N/M and the block had only traveled a bit more than 5 meters at this point. The final velocity of the block that I calculated correctly (finally) was 10.727 meters. The block's mass was 300kg which led me to believe that the spring's compression might be more?

I get a much larger distance. The low spring constant should lead a large compression distance. Please post your working.

 

[spacestrudel]

I get a much larger distance. The low spring constant should lead a large compression distance. Please post your working.

Ah, I see.
Here is my work:

1/2mv^2_f +mgh_f = u_kmgcos(Θ)(x+s) + 1/2ks^2

By plugging in the values I have the following:

1/2(300)(10.727)^2 + (300)(9.81)[-((s+5)sin(30))] = (0.4)(300)(9.81)cos(30)(s+5) + 10s^2
(where 300 = the mass, 10.72 = the final velocity, 9.81 for gravity, sin(30) = 1/2. 0.4 = uk, 300 = mass and 10 because 1/2 of 20 = 10)

Which then leads to:

17260.279 - 1471.5S - 7357.5 = 1019.485S + 5097.426 + 10S^2
This is just 1/2(300)(10.727)^2. I got -1471.5 by multiplying (300)*(9.81)*(-1/2) <--- 1/2 for sin(30) and then times S. The value -7357.5 comes from multiplying (300)*(9.81)*(-1/2)*(5) <--- 5 was part of hf (the distance). Also, they are negative since we said hf was negative.
The RHS of the equation I did the same type of multiplication (0.4) <-- coefficient of friction times (300)*(9.81)*(cos(30))*s = 1019.485S. Then, same thing, this time multiply by 5 instead of s to yield 5097.426.
Moving the terms and simplifying I get the following:

4805.353 - 2490.985S = 10S^2

Which then becomes: 10S^2 + 2490.985S - 4805.353 = 0

Plug in these values into the quadratic formula:

[((-2490.985) + sqrt((2490.985)^2 - 4(10)(-4805.353))]/20 = 1.914 meters

I'm assuming I made an arithmetic mistake somewhere

 

[haruspex]

1/2(300)(10.727)^2

From that I gather you got 10.727m/s for the first part. I get a value over 11.

ukmgcos(Θ)(x+s)

Since you are using the speed at which the mass hits the spring as your initial speed, the work done against friction in reaching the spring is already accounted for. You no longer care about the 5m.

 

[spacestrudel]

From that I gather you got 10.727m/s for the first part. I get a value over 11.

Since you are using the speed at which the mass hits the spring as your initial speed, the work done against friction in reaching the spring is already accounted for. You no longer care about the 5m.

Okay ! I see - thank you.

I used the following equation for calculating the final velocity:

v^2_i +2gxsinΘ -2u_kxcos(Θ) = v^2_f

The 2 comes from the fact that I multiplied everything by 2 to get rid of the 1/2's. Also since everything had "m" on it, I also canceled out all of the m's.

Then I square-rooted both sides and plugged in the values. I believe it must just be a calculation error(?) on my part.