Law of Conservation of Energy Problem: Trampoline

[orangegalaxies]

Homework Statement

Harry has a mass of 58kg and jumps off a roof that is 3.8 m above the ground and lands on a trampoline that is 1 m off the ground. The trampoline has 110 springs each with a k value of 77.6 N/m.

a) How far does the spring/trampoline stretch? For simplicity sake, assume the trampoline mesh is an extension of the springs.
b) What is the acceleration of Harry on the trampoline when his speed reaches 0 m/s? include a free body diagram.
c) If 30% of the energy is lost due to friction each bounce, how far above the trampoline will Harry reach on his first bounce? 2nd bounce?

Relevant Equations

Eg = mgh
Ee = 1/2kx^2
Ek = 1/2mv^2
F = ma
Fg = mg

For a) I did Eg = Ee + Eg and tried to solve for x. I got 5.4 m but I think this is wrong. I have no idea how to do the rest, please help :')

 

[gneill]

Kindly show the details of your calculation attempt. Your answer of 5.4 m is clearly incorrect as the trampoline surface is only 1 m above the ground.

 

[orangegalaxies]

Kindly show the details of your calculation attempt. Your answer of 5.4 m is clearly incorrect as the trampoline surface is only 1 m above the ground.

Eg = Ee + Eg
mgh = 1/2kx^2 + mgh
58(9.8)(3.8) = 1/2(110)x^2 + 58(9.8)(1)
2159.92 = 55x^2 + 568.4
x = 5.4 m

 

[gneill]

Eg = Ee + Eg
mgh = 1/2kx^2 + mgh
58(9.8)(3.8) = 1/2(110)x^2 + 58(9.8)(1)
2159.92 = 55x^2 + 568.4
x = 5.4 m

In your second line mgh cancels on each side and you're left with

0 = 1/2kx^2

That can't be right, as it implies that the trampoline does not stretch at all.

I think you need to be a bit more careful with your variable definitions so that people can follow your equations.

The rest of your calculation, where you've just plugged in numbers, doesn't make sense to me. How can you justify plugging in "110" for the net spring constant? One spring has a given spring constant, so what do you get when you put 110 of them in parallel?

What does the "1" mean in the term 58(9.8)(1)? Are you implying that the trampoline must stretch by 1 m?

Perhaps you could explain your calculation thought process line by line.

 

[orangegalaxies]

In your second line mgh cancels on each side and you're left with

0 = 1/2kx^2

That can't be right, as it implies that the trampoline does not stretch at all.

I think you need to be a bit more careful with your variable definitions so that people can follow your equations.

The rest of your calculation, where you've just plugged in numbers, doesn't make sense to me. How can you justify plugging in "110" for the net spring constant? One spring has a given spring constant, so what do you get when you put 110 of them in parallel?

What does the "1" mean in the term 58(9.8)(1)? Are you implying that the trampoline must stretch by 1 m?

Perhaps you could explain your calculation thought process line by line.

hi, thanks so much. I'm really sorry about the confusion, i made a typo so i will fix it. i tried the question again and got this so i hope it makes more sense.

h1 = 3.8m (guy standing above cliff)
v1 = 0
x1 = 0

h2 = 1m (trampoline)
v2 = 0
x2 = ?

Eg = Eg + Ee (law of conservation)
mgh1 = 1/2kx2^2 + mgh2 (formulas for each of the above based on the variable list above)
58(9.8)(3.8) = (110)1/2(77.6)x^2 + 58(9.8)(1) (subbing in variables. there are 110 springs so i multiplied the Ee equation by 110)
2159.92 = 4268x^2 + 568.4
x = 0.6m

i would really appreciate it if you could tell me if I'm doing this right, and how i can approach parts b and c.

 

[gneill]

You're close, but I think you would benefit by drawing a diagram of what's happening so that you can visualize your energy changes when writing out your equations. For example, before Harry jumps the situation is like this:

Now clearly, Harry falls a distance (h1 - h2) before hitting the trampoline. When the trampoline has maximally stretched the situation is as follows:

As you can see, Harry falls an additional distance Δx, stretching the springs. So that's an additional mgΔx of potential energy going into system. What's the total distance that Harry falls? That gives you the total PE "sourced" and converted to other energy. Can you write an expression that describes the total gravitational PE that goes towards stretching the trampoline springs to their maximum for this situation?

How far do the springs stretch? That gives you the energy transferred to the springs, which should be equal to the "sourced" energy from gravitation.

A nit-pick: In a real-life trampoline the springs don't stretch vertically. They go from approximately horizontal to some angle below the horizontal, so the vertical deflection of the trampoline mat is something different from the actual stretching of the springs depending upon the geometry of the trampoline. It's not a trivial exercise to work out how the energy will be distributed amongst the springs for a non-circular trampoline.

For part (b) you need to find the net force acting on Harry when the trampoline is maximally stretched. If you found Δx while doing part (a) that shouldn't be a problem.

For part (c) you need to consider what losing 30% of the system energy does to the height of the bounce. If he leaves the trampoline surface with 70% of the energy he arrived with, how does that affect height he returns to? Look at how KE and gravitational PE exchange. Also note that they ask for his height above the trampoline, not the ground for this part.

 

[orangegalaxies]

You're close, but I think you would benefit by drawing a diagram of what's happening so that you can visualize your energy changes when writing out your equations. For example, before Harry jumps the situation is like this:
View attachment 275881

Now clearly, Harry falls a distance (h1 - h2) before hitting the trampoline. When the trampoline has maximally stretched the situation is as follows:
View attachment 275882
As you can see, Harry falls an additional distance Δx, stretching the springs. So that's an additional mgΔx of potential energy going into system. What's the total distance that Harry falls? That gives you the total PE "sourced" and converted to other energy. Can you write an expression that describes the total gravitational PE that goes towards stretching the trampoline springs to their maximum for this situation?

How far do the springs stretch? That gives you the energy transferred to the springs, which should be equal to the "sourced" energy from gravitation.

A nit-pick: In a real-life trampoline the springs don't stretch vertically. They go from approximately horizontal to some angle below the horizontal, so the vertical deflection of the trampoline mat is something different from the actual stretching of the springs depending upon the geometry of the trampoline. It's not a trivial exercise to work out how the energy will be distributed amongst the springs for a non-circular trampoline.

For part (b) you need to find the net force acting on Harry when the trampoline is maximally stretched. If you found Δx while doing part (a) that shouldn't be a problem.

For part (c) you need to consider what losing 30% of the system energy does to the height of the bounce. If he leaves the trampoline surface with 70% of the energy he arrived with, how does that affect height he returns to? Look at how KE and gravitational PE exchange. Also note that they ask for his height above the trampoline, not the ground for this part.

ahhhhh, i see! i think this would make sense then:

Eg = Ee
mgh = 1/2kx^2
58(9.8)((3.8-1)+ x) = 1/2(77.6)x^2
1591.53 + 568.4x = 38.8x^2

i would then have to use the quadratic formula to solve for x. i got -0.28m and 15m. 15m seems unreasonable, so do i still have to multiple the Ee formula by the number of springs in the trampoline?

thank you so much for your help! i will attempt b and c and let you know how it goes

edit: i multiplied Ee by the number of springs and got x = 0.7 which seems more sensible but i'll wait for your input

 

[haruspex]

Your answer of 5.4 m is clearly incorrect as the trampoline surface is only 1 m above the ground.

Only a problem if you oversimplify, as below:

A nit-pick: In a real-life trampoline the springs don't stretch vertically. They go from approximately horizontal to some angle below the horizontal, so the vertical deflection of the trampoline mat is something different from the actual stretching of the springs depending upon the geometry of the trampoline. It's not a trivial exercise to work out how the energy will be distributed amongst the springs for a non-circular trampoline.

I wouldn’t consider the direction of stretching a nitpick, but I agree the student is not expected to worry about uneven stretch. Just treat the trampoline as a single horizontal elastic line.
Admittedly, this does lead to a quartic, but maybe we can assume the depression is small compared with the trampoline width.
But therein lies another difficulty, or resolution: we are not told the relaxed length of the "spring". Real trampolines are fairly taut even when nearly flat. It might be reasonable to take the relaxed length as zero, in which case the quartic goes away.
In between we have the worst of all worlds: the relaxed length neither short enough to be ignored nor long enough to dwarf the depression.

Edit: just realized we are not given the width of the trampoline, so the question is a nonsense. We have no choice but to redesign the trampoline as a rigid horizontal plate supported by vertical springs!

 

[kuruman]

What is the justification for using mechanical energy conservation to find the maximum compression when 30% of the energy is lost to friction? Of course to do part (a) we can pretend that this friction does not exist since we learn about it later in part (c). However if that is the case, then when we reach part (c) we shall have to recalculate the maximum compression taking friction into account this time.

 

[haruspex]

when we reach part (c) we shall have to recalculate the maximum compression taking friction into account this time.

I see no need for that. We can forget that it's a trampoline and just say that each time the highest point is reached there's 30% less mechanical energy available.

Edit: on reflection, we do need to find the depression of the trampoline, since that is all part of the lost GPE of which only 70% will be restored. See post #15.

 

[orangegalaxies]

You're close, but I think you would benefit by drawing a diagram of what's happening so that you can visualize your energy changes when writing out your equations. For example, before Harry jumps the situation is like this:
View attachment 275881

Now clearly, Harry falls a distance (h1 - h2) before hitting the trampoline. When the trampoline has maximally stretched the situation is as follows:
View attachment 275882
As you can see, Harry falls an additional distance Δx, stretching the springs. So that's an additional mgΔx of potential energy going into system. What's the total distance that Harry falls? That gives you the total PE "sourced" and converted to other energy. Can you write an expression that describes the total gravitational PE that goes towards stretching the trampoline springs to their maximum for this situation?

How far do the springs stretch? That gives you the energy transferred to the springs, which should be equal to the "sourced" energy from gravitation.

A nit-pick: In a real-life trampoline the springs don't stretch vertically. They go from approximately horizontal to some angle below the horizontal, so the vertical deflection of the trampoline mat is something different from the actual stretching of the springs depending upon the geometry of the trampoline. It's not a trivial exercise to work out how the energy will be distributed amongst the springs for a non-circular trampoline.

For part (b) you need to find the net force acting on Harry when the trampoline is maximally stretched. If you found Δx while doing part (a) that shouldn't be a problem.

For part (c) you need to consider what losing 30% of the system energy does to the height of the bounce. If he leaves the trampoline surface with 70% of the energy he arrived with, how does that affect height he returns to? Look at how KE and gravitational PE exchange. Also note that they ask for his height above the trampoline, not the ground for this part.

hi again! for a) i did what i mentioned above and got 0.7m.

for b), i equated Fe + Fg = ma.
kx + mg = ma
110(11.6)(0.7) + 58(9.8) = 58a
a = 93 m/s^2

for c), i found Ee.
1/2kx^2
1/2(110)(77.6)(0.7)^2 = 2091 J
then i found 70% of it for the first jump = 1464 J
i equated this to mgh
1464 = mgh
h = 2.6m

i did the same for the second jump, after finding 70% of 1464J

am i on the right track?

 

[kuruman]

I see no need for that. We can forget that it's a trampoline and just say that each time the highest point is reached there's 30% less mechanical energy available.

I agree with that. My concern is that the use of mechanical energy conservation to find the answer in part (a) when the system is dissipative might send a conflicting message to someone asking oneself "what did solving this problem teach me"?

 

[orangegalaxies]

What is the justification for using mechanical energy conservation to find the maximum compression when 30% of the energy is lost to friction? Of course to do part (a) we can pretend that this friction does not exist since we learn about it later in part (c). However if that is the case, then when we reach part (c) we shall have to recalculate the maximum compression taking friction into account this time.

aah i don't know dude I'm still in high school :'))

 

[haruspex]

First, please see my edit at the end of post #8.

hi again! for a) i did what i mentioned above and got 0.7m.

looks right, except that you would be justified in quoting one more digit of precision.

for b), i equated Fe + Fg = ma.
kx + mg = ma

Which way are the spring force and gravitational force acting?

110(11.6)(0.7) + 58(9.8) = 58a

Where does the 11.6 come from?
Also, wherever you have numeric values plugged in you should quote the units.

for c),
1/2(110)(77.6)(0.7)^2 = 2091 J

By plugging in the rounded value of 0.7 you are losing precision. Use three sig figs at least.

then i found 70% of it for the first jump = 1464 J
i equated this to mgh
1464 = mgh

Where h is what, exactly? Be careful here.

 

[haruspex]

What is the justification for using mechanical energy conservation to find the maximum compression when 30% of the energy is lost to friction? Of course to do part (a) we can pretend that this friction does not exist since we learn about it later in part (c). However if that is the case, then when we reach part (c) we shall have to recalculate the maximum compression taking friction into account this time.

Corrected response:
It isn't friction, in the usual sense. It is that on rebound the spring constant will be only 70% of the given 110N/m. All the work in parts a and b is still valid.

 

[kuruman]

Corrected response:
It isn't friction, in the usual sense. It is that on rebound the spring constant will be only 70% of the given 110N/m. All the work in parts a and b is still valid.

So the model is that while Harry is in contact with the trampoline, no energy is lost when his velocity is down, but once he reverses direction 30% is dissipated by the time he loses contact going back up.

 

[haruspex]

So the model is that while Harry is in contact with the trampoline, no energy is lost when his velocity is down, but once he reverses direction 30% is dissipated by the time he loses contact going back up.

I wouldn't put it that way.
A spring loses energy through hysteresis, its constant being greater during active stretching (or compression) than during relaxation. It is not possible (?) to apportion the loss between the two parts of the cycle.
In order to solve the first part of the question, we have to assume the given constant applies to the stretching phase.