Practice Problem: Launching Things With Springs
ฝัง
- เผยแพร่เมื่อ 28 ก.ย. 2024
- You're a mischievous little physicist, aren't you? Well I guess there's no harm in projecting a toy into the air, especially if it's part of your scientific research. Given a spring constant and some other information, how high will this little army guy go? Let's find out!
Try all of the Classical Physics Practice Problems: bit.ly/ProfDave...
Classical Physics Tutorials: bit.ly/ProfDave...
Modern Physics Tutorials: bit.ly/ProfDave...
Mathematics Tutorials: bit.ly/ProfDave...
General Chemistry Tutorials: bit.ly/ProfDave...
Organic Chemistry Tutorials: bit.ly/ProfDave...
Biochemistry Tutorials: bit.ly/ProfDave...
Biology Tutorials: bit.ly/ProfDaveBio
EMAIL► ProfessorDaveExplains@gmail.com
PATREON► / professordaveexplains
Check out "Is This Wi-Fi Organic?", my book on disarming pseudoscience!
Amazon: amzn.to/2HtNpVH
Bookshop: bit.ly/39cKADM
Barnes and Noble: bit.ly/3pUjmrn
Book Depository: bit.ly/3aOVDlT
Wrong. force is not constant while the sprig is relaxed. the way we can solve this problem is by applying the conservation of energy principle at the lowest point where the spring compressed (storing elastic energy with zero kinetic and considering 0 potential) and at the point where it is relaxed (having kinetic and potential energies and zero elastic) solving this we can know initial velocity by which the person is ejected upwards. then we easily apply motion under the gravitational acc. equation for finding any other values.
but sir
I think that that acceleration will not be constant
that is, it would continuously decrease as the spring expands
so it must've been done by integration or better conservation of energy
please guide and reply🤔🤔🤔🤔🤔🤔
but still it was an interesting Question😊😊👌👍
Sorry for replying to an old comment but I'm really interested in how would this be solved without using conservation of energy? I'm just getting into calculus and I can't figure it out.
I think you are imagining an instant application of the force by the spring, but it is applied throughout the time when the spring is expanding so the acceleration should be constant.
You should imagine force is just the change of momentum, so at every instant the velocity will change accordingly
@@alwaysy5178 If we consider the hit to be perfectly elastic, it doesn't matter if spring pushes from the beginning or hits the object positioned at length of uncompressed spring, it's effectively the same. Force is applied throughout the time but it is not constant. It changes depending on the displacement at given moment, so how could acceleration then possibly be constant. Think about it, the more you want to compress the spring, the more force you need to apply and the more spring is compressed, the harder it pushes back. It pushes back with gradually decreasing force and with it, proportionally decreasing acceleration (F=ma). Calculating the velocity at 0 displacement without the law of conservation of energy requires harmonic oscillator calculus.
@@mani225456 Condition for spring's SHM is that its displacement x is a cos graph (starts at x₀ - derivation that x is cos graph is a linear ODE or some trigonometric intuition with circular motion). Specifically the cos of θ (where θ = 2π is a full oscillation), and to adjust the cos graph for its amplitude x₀ the result is x = x₀ cos θ. To obtain its 'launch' velocity, we must determine its velocity at equilibrium or x = 0. v = dx/dt = x'(t); we can obtain a value to determine x with respect to t instead of with respect to θ using the frequency of the oscillation ω which represents the radians of oscillation per unit of time. Hence θ can be rewritten as ωt, so x = x₀ cos ωt. Derivative of x with respect to t is then (using chain rule) d/dt x₀ cos ωt = -ωx₀ sin ωt. Now solve for velocity at equilibrium x = 0.
First we need to find this spring's angular frequency ω. We know the spring obeys Hooke's Law, therefore F = -kx -> ma = -kx -> a = -kx/m. We need to find the acceleration of the spring with respect to t which is the derivative of velocity with respect to t = d/dt -ωx₀ sin ωt which, again using chain rule, yields a(t) = v'(t) = x''(t) = -ω²x₀ cos ωt. -kx/m = -ω²x₀ cos ωt; noticing that "x₀ cos ωt" is simply x with respect to t we can substitute -kx/m = -ω²x -> k/m = ω² -> ω = √(k/m). Finally we can put some numbers to this: ω = √(85.4/0.01) ≈ 92.4, so around 92 radians per second.
v at equilibrium, we know ωt = θ which is π/2 at equilibrium. Hence v = -ωx₀ sin π/2 = (-92.4)(-0.04) sin π/2 ≈ 3.70 m/s. Finally we can just use suvat equations or I guess if you really want integrate ∫ -9.81t + 3.70 from t = 0 and (using v = u + at -> 0 = 3.70 - 9.81t -> t = 0.377, as v will be 0 at the peak of its launch) t = 0.377 (x-intercept) yields approximately 0.698 meters (depends on how you rounded) above the launch point which is at 10 cm so the maximum height, final answer, is approximately 0.798 m or 79.8 cm. (not what the video has obviously as it was a completely wrong method)
Work is measured in Joules and Force is measured in Newtons. So wouldn't the "Net" Force be different then what you have since you need to convert the Work to Force from the resistance of gravity on the object?
P.S. I like your videos, typically very top-notch!
Using Conservation of Energy, you get the energy stored in the spring when compressed 0.040 m is 0.06832 Joules. When converting this to Gravitational Potential Energy, you can calculate the height to be 0.697 m. This is measured from the lowest point, when the spring is compressed, so it will rise 0.657 m above the uncompressed spring. Since the force to compress a spring changes linearly as the compression distance (hooke's law), the force is not constant, therefore you cannot assume a constant acceleration at during the launch. Elastic Energy -> Gravitational Potential Energy.
Sir, why would it be wrong to directly apply energy conservation to this problem, also If we solve it by energy conservation should we use 1/2kx^2 or kx, despite being in 12th I do get hopelessly confused with that small step. Also if we do use energy conservation we do not have to consider mg downwards right?
hmm, energy conservation in what way? like how would you set up the equation? feel free to email me as well.
its a bit silly but 1/2kx^2=1/2mv^2, we solve for v then plug it into the formula for max height v^2/2g (its basically derived from v^2 - u^2=2AS which is what you used as well)
hey if you get the right answer, then it sounds good to me!
I'm getting half of the answer you got.....
My result was 0.7946m.
First, F(x) = -2135x+85.4 is the force in function of distance, where it reaches 85,4 when compressed (x = 0) and 0 when relaxed (x = 0.04).
Then, R(x) = F(x) - 9.8 * 0.01 is the resulting force in function of distance.
A(x) = R(x)/0.01 is acceleration in function of distance.
As A(x) = 0 when x = 0.03995, the integral of A(x) with respect of x, in the interval of (0, 0.03995) will equal to 170.048, wich can be interpreted as the mean acceleration during the spring relaxation part.
Using 1/2 * at^2 = d to figure out the time passed during the relaxation give us t = sqrt(0.03995 * 2 / (170.048)) = 0.0216.
Then, t * a should give the velocity at the end of the relaxation: 3.689 m/s
-9.8 m/s^2 * t + 3.689 m/s = 0 when t = 0.3765 s, is the time at wich the soldier gets its peak height whith that initial velocity.
4.9 m/s^2 * t^2 + 0.09995m = 0.7946m, wich should be the top height, where 0.09995m is the height of the spring, not considering that, it would be 0.6946m.
This video is not correct so be careful watching it. Force & acceleration are not constant at all for a spring. You’d have to use energy conservation, not kinematics. You over compensate by an entire factor of 2 when using the videos technique. Not even close.
I used the formula 1/2kxmax^2 = mghmax and I got 0.68 meters. How come this method didn’t work? Thanks, love your content 👨❤️💋👨
Nah, your method is the correct way. In the video he incorrectly assumed the force on the toy during the expansion of the spring was constant throughout. Remember F=-kx, so that means at the beginning of the expansion the force was the strongest and as the spring relaxes the force gets weaker.
TrippyTurtle exactly! I got to 0,6832 meters as well Standard Trevor
@@Trippy_Turtle but for (1/2).k.xmax^2 = m.g.hmax to be true all the energy from the spring would need to be transfered to the army guy, right?
@@JojonathanOliveira The Xmax is how much the spring was pushed down, and that determines how much energy the spring gives to the army man. Where else would the energy of the spring go to? The ground is too massive to move at all.
@@Trippy_Turtle wouldn't the spring bounce around a little after the weight(army man) is ejected? It needs energy to do that, I think it would eventually lose it to friction and air resistance after a while
Sir I always saw u as a perfect professor. And u r the reason why I got some publicity due to science in my class
Sir i do have a small request for you . Please tell me video link of how to protect alcohol with tosyl chloride
i did that already! i have a whole clip on protecting groups. and please keep comments relevant to the clip you are commenting on, if you have a general question or request, just email me.
I don't know if this is explained or not, but why would the spring only go 4 centimeters after being compressed for the same length. How can the spring be made so it compresses only a little bit but extends alot on release?
Hi, the distance is 1.35m. so let’s say if we add 3 more extra same springs and bounce the same figurine, would the distance be multiplied by 4 ?
great video I learned quite a bit.
Superb video sir. 👍👍👍👍👍👍
Awesome
Sir max height achieved also depends upon the shape of the object.
not if we disregard wind resistance, which we tend to do in introductory physics.
Sir where are this concepts application?
all kinds of stuff! kinematics applies to all objects in motion. springs demonstrate newton's laws. plus springs are neat in their own right.