An error made in a kinematics question about a spring launcher?

In summary, the conversation discusses the Nifty Numbat Launcher, which is created by Genevieve the galah. The launcher consists of a large spring with an equilibrium length of 1.7 m and a spring constant of 41 N/m. The spring is attached to a ramp inclined at a 30 degree angle to the horizontal. Genevieve compresses the spring to a length of 0.4 m and her mate Nigel, who has a mass of 2.5 kg, hops onto the ramp. When the spring is released, Nigel will be launched into the air. The group discusses the speed at which Nigel will be launched and different equations and methods are suggested, with a final consensus of 3.87
  • #1
148
5
Homework Statement
below
Relevant Equations
v^2=u^2+2as
Genevieve the galah wants to test out her new creation - the Nifty Numbat Launcher. It consists of a large, ideal spring with equilibrium length l1 = 1.7 m and spring constant k=41 N/m. The spring rests upon a ramp of length l1 inclined at an angle β=30∘ to the horizontal. The spring is attached to the bottom of the ramp. Genevieve compresses the spring to a shorter length l2 = 0.4 m, and convinces her mate Nigel the numbat (mass m = 2.5 kg) to hop onto the ramp in front of the spring. When she releases the compressed spring, Nigel will be launched into the air! But just how fast will he be launched?

Below is a diagram of the Nifty Numbat Launcher in action.

1594114221121.png

I am getting 3.87 for velocity. Answer says 3.325. No idea how they got it.

I drew a basic force diagram, spring force upwards 41x1.3/2 - gravity component force downwards 9.8x2.5sin30. Net force/2.5. Then use v^2=u^2+2as.
 
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  • #2
The equation you are using is meant to be used when the acceleration is constant.
Do you remember the equation for the elastic force? Is it constant?
 
  • #3
archaic said:
The equation you are using is meant to be used when the acceleration is constant.
Do you remember the equation for the elastic force? Is it constant?

I took an average acceleration which basically works as a constant acceleration. i.e. spring force at the start is at x1.3m, at the end it's 0, the equivalent constant value is the average of these two values. Does it not?
 
  • #4
aspodkfpo said:
I took an average acceleration which basically works as a constant acceleration. i.e. spring force at the start is at x1.3m, at the end it's 0, the equivalent constant value is the average of these two values. Does it not?
No!
 
  • #5
PeroK said:
No!

What, how does this work then?
 
  • #7
aspodkfpo said:
What, how does this work then?
Using energy is much better. In general, if you want to use forces, you must integrate. Constant acceleration is just a special case where you have the SUVAT formulae.
 
  • #9
aspodkfpo said:
F=kx, how does averaging the ends not give me an equivalent constant force?
That's an average over distance. You converted that to an average over time. The time profile of the force is harmonic, not linear.
 
  • #10
PeroK said:
That's an average over distance. You converted that to an average over time. The time profile of the force is harmonic, not linear.

Oh right, because it's accelerating it can't be an average over time. Correct me if I'm wrong.
 
  • #11
aspodkfpo said:
Oh right, because it's accelerating it can't be an average over time. Correct me if I'm wrong.
It spends longer time wise at the greater force, so yes it's not a time-based average if it's distance based.
 
  • #12
archaic said:
The equation you are using is meant to be used when the acceleration is constant.
True, but it also works if the acceleration used is the average over distance (which, to be clear, is not the "average acceleration", which always means average over time).
Average acceleration over distance =##\frac{\int a.ds}{\int ds}=\frac{\int ma.ds}{m\Delta s}=\frac{\Delta E}{m\Delta s}=\frac{\frac 12 (v^2-u^2)}{\Delta s}##
PeroK said:
Using energy is much better.
In effect, that is what @aspodkfpo did.
PeroK said:
That's an average over distance. You converted that to an average over time.
I don't see any such conversion.

Using work, I also get 3.87 m/s. Anybody see a way to get 3.325?
 
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  • #13
haruspex said:
True, but it also works if the acceleration used is the average over distance (which, to be clear, is not the "average acceleration", which always means average over time).
Average acceleration over distance =##\frac{\int a.ds}{\int ds}=\frac{\int ma.ds}{m\Delta s}=\frac{\Delta E}{m\Delta s}=\frac{\frac 12 (v^2-u^2)}{\Delta s}##

In effect, that is what @aspodkfpo did.

I don't see any such conversion.

Using work, I also get 3.87 m/s. Anybody see a way to get 3.325?

It's a good point. Generally, though, using SUVAT formulae outside of constant acceleration is asking for trouble. It was lucky in this case that the formula used was valid.
 
  • #14
haruspex said:
True, but it also works if the acceleration used is the average over distance (which, to be clear, is not the "average acceleration", which always means average over time).
Average acceleration over distance =##\frac{\int a.ds}{\int ds}=\frac{\int ma.ds}{m\Delta s}=\frac{\Delta E}{m\Delta s}=\frac{\frac 12 (v^2-u^2)}{\Delta s}##

In effect, that is what @aspodkfpo did.

I don't see any such conversion.

Using work, I also get 3.87 m/s. Anybody see a way to get 3.325?

Still getting 3.87 with work as well. In what scenarios am I allowed to use average acceleration over distance? Only for E=Fs right? And it would be wrong if I were to use average acceleration over time for that formula since I am multiplying it by distance? I'm getting confused here.
 
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  • #15
aspodkfpo said:
Still getting 3.87 with work as well. In what scenarios am I allowed to use average acceleration over distance? Only for E=Fs right? And it would be wrong if I were to use average acceleration over time for that formula since I am multiplying it by distance? I'm getting confused here.
Not sure you realize I confirm that the given answer is wrong.

E=Fs (or, cancelling m, ##v^2-u^2=2a_{average-over-distance}##) you can take as the definition of average acceleration over distance. So wrt using it, the interesting question is when can you find that average by other means.
In the present case you could because you knew the force changes linearly with distance. Constant force may be the only other case where it is easy to find it.
 
  • #16
haruspex said:
Not sure you realize I confirm that the given answer is wrong.

E=Fs (or, cancelling m, ##v^2-u^2=2a_{average-over-distance}##) you can take as the definition of average acceleration over distance. So wrt using it, the interesting question is when can you find that average by other means.
In the present case you could because you knew the force changes linearly with distance. Constant force may be the only other case where it is easy to find it.

"
Not sure you realize I confirm that the given answer is wrong.
"

Just confirming the fact that I'm getting the same answer as you with the work method, so that maybe someone else can see another (probably wrong) method of getting 3.325.
 
  • #17
aspodkfpo said:
"
Not sure you realize I confirm that the given answer is wrong.
"

Just confirming the fact that I'm getting the same answer as you with the work method, so that maybe someone else can see another (probably wrong) method of getting 3.325.
I tried a couple of blunders, but could not get that number.
 
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